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MA 108 - Ordinary Differential Equations
Santanu Dey
Department of Mathematics,Indian Institute of Technology Bombay,
Powai, Mumbai [email protected]
February 25, 2014
Santanu Dey Lecture 2
Outline of the lecture
1 Initial value problems
2 Separable DE
3 Equations reducible to separable form
Santanu Dey Lecture 2
First order ODE & Initial Value Problem for first orderODE
We now consider first order ODE of the form F (x , y , y ′) = 0 or
y ′ = f (x , y) .
Consider a linear first order ODE of the form y ′ + a(x)y = b(x) .
If b(x) = 0, then we say that the equation is homogeneous.
Note that the solutions of a homogeneous differential equationform a vector space under usual addition and scalar multiplication
Definition
Initial value problem (IVP) : A DE along with an initial condition isan IVP.
y ′ = f (x , y), y(x0) = y0.
Santanu Dey Lecture 2
First order ODE & Initial Value Problem for first orderODE
We now consider first order ODE of the form F (x , y , y ′) = 0 or
y ′ = f (x , y) .
Consider a linear first order ODE of the form y ′ + a(x)y = b(x) .
If b(x) = 0, then we say that the equation is homogeneous.
Note that the solutions of a homogeneous differential equationform a vector space under usual addition and scalar multiplication
Definition
Initial value problem (IVP) : A DE along with an initial condition isan IVP.
y ′ = f (x , y), y(x0) = y0.
Santanu Dey Lecture 2
First order ODE & Initial Value Problem for first orderODE
We now consider first order ODE of the form F (x , y , y ′) = 0 or
y ′ = f (x , y) .
Consider a linear first order ODE of the form y ′ + a(x)y = b(x) .
If b(x) = 0, then we say that the equation is homogeneous.
Note that the solutions of a homogeneous differential equationform a vector space under usual addition and scalar multiplication
Definition
Initial value problem (IVP) : A DE along with an initial condition isan IVP.
y ′ = f (x , y), y(x0) = y0.
Santanu Dey Lecture 2
First order ODE & Initial Value Problem for first orderODE
We now consider first order ODE of the form F (x , y , y ′) = 0 or
y ′ = f (x , y) .
Consider a linear first order ODE of the form y ′ + a(x)y = b(x) .
If b(x) = 0, then we say that the equation is homogeneous.
Note that the solutions of a homogeneous differential equationform a vector space under usual addition and scalar multiplication
Definition
Initial value problem (IVP) : A DE along with an initial condition isan IVP.
y ′ = f (x , y), y(x0) = y0.
Santanu Dey Lecture 2
First order ODE & Initial Value Problem for first orderODE
We now consider first order ODE of the form F (x , y , y ′) = 0 or
y ′ = f (x , y) .
Consider a linear first order ODE of the form y ′ + a(x)y = b(x) .
If b(x) = 0, then we say that the equation is homogeneous.
Note that the solutions of a homogeneous differential equationform a vector space under usual addition and scalar multiplication
Definition
Initial value problem (IVP) : A DE along with an initial condition isan IVP.
y ′ = f (x , y), y(x0) = y0.
Santanu Dey Lecture 2
Examples
Given an amount of a radioactive substance, say 1 gm, find theamount present at any later time.
The relevant ODE isdy
dt= −k · y .
Initial amount given is 1 gm at time t = 0. i.e.,
y(0) = 1.
By inspection, y = ce−kt , for an arbitrary constant c , is a solutionof the above ODE. The initial condition determines c = 1. Hence
y = e−kt
is a particular solution to the above ODE with the given initialcondition.
Santanu Dey Lecture 2
Examples
Given an amount of a radioactive substance, say 1 gm, find theamount present at any later time.The relevant ODE is
dy
dt= −k · y .
Initial amount given is 1 gm at time t = 0. i.e.,
y(0) = 1.
By inspection, y = ce−kt , for an arbitrary constant c , is a solutionof the above ODE. The initial condition determines c = 1. Hence
y = e−kt
is a particular solution to the above ODE with the given initialcondition.
Santanu Dey Lecture 2
Examples
Given an amount of a radioactive substance, say 1 gm, find theamount present at any later time.The relevant ODE is
dy
dt= −k · y .
Initial amount given is 1 gm at time t = 0.
i.e.,
y(0) = 1.
By inspection, y = ce−kt , for an arbitrary constant c , is a solutionof the above ODE. The initial condition determines c = 1. Hence
y = e−kt
is a particular solution to the above ODE with the given initialcondition.
Santanu Dey Lecture 2
Examples
Given an amount of a radioactive substance, say 1 gm, find theamount present at any later time.The relevant ODE is
dy
dt= −k · y .
Initial amount given is 1 gm at time t = 0. i.e.,
y(0) = 1.
By inspection, y = ce−kt , for an arbitrary constant c , is a solutionof the above ODE. The initial condition determines c = 1. Hence
y = e−kt
is a particular solution to the above ODE with the given initialcondition.
Santanu Dey Lecture 2
Examples
Given an amount of a radioactive substance, say 1 gm, find theamount present at any later time.The relevant ODE is
dy
dt= −k · y .
Initial amount given is 1 gm at time t = 0. i.e.,
y(0) = 1.
By inspection, y = ce−kt , for an arbitrary constant c , is a solutionof the above ODE.
The initial condition determines c = 1. Hence
y = e−kt
is a particular solution to the above ODE with the given initialcondition.
Santanu Dey Lecture 2
Examples
Given an amount of a radioactive substance, say 1 gm, find theamount present at any later time.The relevant ODE is
dy
dt= −k · y .
Initial amount given is 1 gm at time t = 0. i.e.,
y(0) = 1.
By inspection, y = ce−kt , for an arbitrary constant c , is a solutionof the above ODE. The initial condition determines c =
1. Hence
y = e−kt
is a particular solution to the above ODE with the given initialcondition.
Santanu Dey Lecture 2
Examples
Given an amount of a radioactive substance, say 1 gm, find theamount present at any later time.The relevant ODE is
dy
dt= −k · y .
Initial amount given is 1 gm at time t = 0. i.e.,
y(0) = 1.
By inspection, y = ce−kt , for an arbitrary constant c , is a solutionof the above ODE. The initial condition determines c = 1. Hence
y = e−kt
is a particular solution to the above ODE with the given initialcondition.
Santanu Dey Lecture 2
Geometrical meaning :dy
dt= −2 · y
1 Suppose that y has certain value.
From the RHS of the DE, we
obtaindy
dt. For instance, if y = 1.5,
dy
dt= −3. This means that the
slope of a solution y = y(t) has the value −3 at any point wherey = 1.5.
2 Display this information graphically in ty -plane by drawing short linesegments or arrows at several points on y = 1.5.
3 Similarly proceed for other values of y .
4 The figures given in next two slides are examples of direction fieldsor slope fields.
5 An isocline (a series of lines with the same slope) is often used tosupplement the slope field. In an equation of the formdy
dt= f (t, y), the isocline is a line in the ty -plane obtained by
setting f (t, y) equal to a constant.
Santanu Dey Lecture 2
Geometrical meaning :dy
dt= −2 · y
1 Suppose that y has certain value. From the RHS of the DE, we
obtaindy
dt.
For instance, if y = 1.5,dy
dt= −3. This means that the
slope of a solution y = y(t) has the value −3 at any point wherey = 1.5.
2 Display this information graphically in ty -plane by drawing short linesegments or arrows at several points on y = 1.5.
3 Similarly proceed for other values of y .
4 The figures given in next two slides are examples of direction fieldsor slope fields.
5 An isocline (a series of lines with the same slope) is often used tosupplement the slope field. In an equation of the formdy
dt= f (t, y), the isocline is a line in the ty -plane obtained by
setting f (t, y) equal to a constant.
Santanu Dey Lecture 2
Geometrical meaning :dy
dt= −2 · y
1 Suppose that y has certain value. From the RHS of the DE, we
obtaindy
dt. For instance, if y = 1.5,
dy
dt= −3. This means that the
slope of a solution y = y(t) has the value −3 at any point wherey = 1.5.
2 Display this information graphically in ty -plane by drawing short linesegments or arrows at several points on y = 1.5.
3 Similarly proceed for other values of y .
4 The figures given in next two slides are examples of direction fieldsor slope fields.
5 An isocline (a series of lines with the same slope) is often used tosupplement the slope field. In an equation of the formdy
dt= f (t, y), the isocline is a line in the ty -plane obtained by
setting f (t, y) equal to a constant.
Santanu Dey Lecture 2
Geometrical meaning :dy
dt= −2 · y
1 Suppose that y has certain value. From the RHS of the DE, we
obtaindy
dt. For instance, if y = 1.5,
dy
dt= −3. This means that the
slope of a solution y = y(t) has the value −3 at any point wherey = 1.5.
2 Display this information graphically in ty -plane by drawing short linesegments or arrows at several points on y = 1.5.
3 Similarly proceed for other values of y .
4 The figures given in next two slides are examples of direction fieldsor slope fields.
5 An isocline (a series of lines with the same slope) is often used tosupplement the slope field. In an equation of the formdy
dt= f (t, y), the isocline is a line in the ty -plane obtained by
setting f (t, y) equal to a constant.
Santanu Dey Lecture 2
Geometrical meaning :dy
dt= −2 · y
1 Suppose that y has certain value. From the RHS of the DE, we
obtaindy
dt. For instance, if y = 1.5,
dy
dt= −3. This means that the
slope of a solution y = y(t) has the value −3 at any point wherey = 1.5.
2 Display this information graphically in ty -plane by drawing short linesegments or arrows at several points on y = 1.5.
3 Similarly proceed for other values of y .
4 The figures given in next two slides are examples of direction fieldsor slope fields.
5 An isocline (a series of lines with the same slope) is often used tosupplement the slope field. In an equation of the formdy
dt= f (t, y), the isocline is a line in the ty -plane obtained by
setting f (t, y) equal to a constant.
Santanu Dey Lecture 2
Geometrical meaning :dy
dt= −2 · y
1 Suppose that y has certain value. From the RHS of the DE, we
obtaindy
dt. For instance, if y = 1.5,
dy
dt= −3. This means that the
slope of a solution y = y(t) has the value −3 at any point wherey = 1.5.
2 Display this information graphically in ty -plane by drawing short linesegments or arrows at several points on y = 1.5.
3 Similarly proceed for other values of y .
4 The figures given in next two slides are examples of direction fieldsor slope fields.
5 An isocline (a series of lines with the same slope) is often used tosupplement the slope field. In an equation of the formdy
dt= f (t, y), the isocline is a line in the ty -plane obtained by
setting f (t, y) equal to a constant.
Santanu Dey Lecture 2
Geometrical meaning :dy
dt= −2 · y
1 Suppose that y has certain value. From the RHS of the DE, we
obtaindy
dt. For instance, if y = 1.5,
dy
dt= −3. This means that the
slope of a solution y = y(t) has the value −3 at any point wherey = 1.5.
2 Display this information graphically in ty -plane by drawing short linesegments or arrows at several points on y = 1.5.
3 Similarly proceed for other values of y .
4 The figures given in next two slides are examples of direction fieldsor slope fields.
5 An isocline (a series of lines with the same slope) is often used tosupplement the slope field. In an equation of the formdy
dt= f (t, y), the isocline is a line in the ty -plane obtained by
setting f (t, y) equal to a constant.
Santanu Dey Lecture 2
Geometrical meaning :dy
dt= −2 · y
1 Suppose that y has certain value. From the RHS of the DE, we
obtaindy
dt. For instance, if y = 1.5,
dy
dt= −3. This means that the
slope of a solution y = y(t) has the value −3 at any point wherey = 1.5.
2 Display this information graphically in ty -plane by drawing short linesegments or arrows at several points on y = 1.5.
3 Similarly proceed for other values of y .
4 The figures given in next two slides are examples of direction fieldsor slope fields.
5 An isocline (a series of lines with the same slope) is often used tosupplement the slope field. In an equation of the formdy
dt= f (t, y), the isocline is a line in the ty -plane obtained by
setting f (t, y) equal to a constant.
Santanu Dey Lecture 2
Slope field
Santanu Dey Lecture 2
Slope field
Santanu Dey Lecture 2
Examples
Find the curve through the point (1, 1) in the xy -plane having at
each of its points, the slope −y
x.
The relevant ODE isy ′ = −y
x.
By inspection,
y =c
xis its general solution for an arbitrary constant c ; that is, a familyof hyperbolas.The initial condition given is
y(1) = 1,
which implies c = 1. Hence the particular solution for the aboveproblem is
y =1
x.
Santanu Dey Lecture 2
Examples
Find the curve through the point (1, 1) in the xy -plane having at
each of its points, the slope −y
x.
The relevant ODE isy ′ = −y
x.
By inspection,
y =c
xis its general solution for an arbitrary constant c ; that is, a familyof hyperbolas.The initial condition given is
y(1) = 1,
which implies c = 1. Hence the particular solution for the aboveproblem is
y =1
x.
Santanu Dey Lecture 2
Examples
Find the curve through the point (1, 1) in the xy -plane having at
each of its points, the slope −y
x.
The relevant ODE isy ′ = −y
x.
By inspection,
y =c
xis its general solution for an arbitrary constant c ; that is, a familyof hyperbolas.
The initial condition given is
y(1) = 1,
which implies c = 1. Hence the particular solution for the aboveproblem is
y =1
x.
Santanu Dey Lecture 2
Examples
Find the curve through the point (1, 1) in the xy -plane having at
each of its points, the slope −y
x.
The relevant ODE isy ′ = −y
x.
By inspection,
y =c
xis its general solution for an arbitrary constant c ; that is, a familyof hyperbolas.The initial condition given is
y(1) = 1,
which implies c = 1.
Hence the particular solution for the aboveproblem is
y =1
x.
Santanu Dey Lecture 2
Examples
Find the curve through the point (1, 1) in the xy -plane having at
each of its points, the slope −y
x.
The relevant ODE isy ′ = −y
x.
By inspection,
y =c
xis its general solution for an arbitrary constant c ; that is, a familyof hyperbolas.The initial condition given is
y(1) = 1,
which implies c = 1. Hence the particular solution for the aboveproblem is
y =1
x.
Santanu Dey Lecture 2
Slope field
Santanu Dey Lecture 2
Slope field
Santanu Dey Lecture 2
Examples
A first order IVP can have
1 NO solution :
|y ′|+ |y | = 0, y(0) = 3.
2 Precisely one solution : y ′ = x , y(0) = 1. What is thesolution?
3 Infinitely many solutions: xy ′ = y − 1, y(0) = 1 The solutionsare y = 1 + cx .
Motivation to study conditions under which the solution wouldexist and the conditions under which it will be unique!
We first start with a few methods for finding out the solution offirst order ODEs, discuss the geometric meaning of solutions andthen proceed to study existence-uniqueness results.
Santanu Dey Lecture 2
Examples
A first order IVP can have
1 NO solution : |y ′|+ |y | = 0, y(0) = 3.
2 Precisely one solution : y ′ = x , y(0) = 1. What is thesolution?
3 Infinitely many solutions: xy ′ = y − 1, y(0) = 1 The solutionsare y = 1 + cx .
Motivation to study conditions under which the solution wouldexist and the conditions under which it will be unique!
We first start with a few methods for finding out the solution offirst order ODEs, discuss the geometric meaning of solutions andthen proceed to study existence-uniqueness results.
Santanu Dey Lecture 2
Examples
A first order IVP can have
1 NO solution : |y ′|+ |y | = 0, y(0) = 3.
2 Precisely one solution :
y ′ = x , y(0) = 1. What is thesolution?
3 Infinitely many solutions: xy ′ = y − 1, y(0) = 1 The solutionsare y = 1 + cx .
Motivation to study conditions under which the solution wouldexist and the conditions under which it will be unique!
We first start with a few methods for finding out the solution offirst order ODEs, discuss the geometric meaning of solutions andthen proceed to study existence-uniqueness results.
Santanu Dey Lecture 2
Examples
A first order IVP can have
1 NO solution : |y ′|+ |y | = 0, y(0) = 3.
2 Precisely one solution : y ′ = x , y(0) = 1.
What is thesolution?
3 Infinitely many solutions: xy ′ = y − 1, y(0) = 1 The solutionsare y = 1 + cx .
Motivation to study conditions under which the solution wouldexist and the conditions under which it will be unique!
We first start with a few methods for finding out the solution offirst order ODEs, discuss the geometric meaning of solutions andthen proceed to study existence-uniqueness results.
Santanu Dey Lecture 2
Examples
A first order IVP can have
1 NO solution : |y ′|+ |y | = 0, y(0) = 3.
2 Precisely one solution : y ′ = x , y(0) = 1. What is thesolution?
3 Infinitely many solutions:
xy ′ = y − 1, y(0) = 1 The solutionsare y = 1 + cx .
Motivation to study conditions under which the solution wouldexist and the conditions under which it will be unique!
We first start with a few methods for finding out the solution offirst order ODEs, discuss the geometric meaning of solutions andthen proceed to study existence-uniqueness results.
Santanu Dey Lecture 2
Examples
A first order IVP can have
1 NO solution : |y ′|+ |y | = 0, y(0) = 3.
2 Precisely one solution : y ′ = x , y(0) = 1. What is thesolution?
3 Infinitely many solutions: xy ′ = y − 1, y(0) = 1
The solutionsare y = 1 + cx .
Motivation to study conditions under which the solution wouldexist and the conditions under which it will be unique!
We first start with a few methods for finding out the solution offirst order ODEs, discuss the geometric meaning of solutions andthen proceed to study existence-uniqueness results.
Santanu Dey Lecture 2
Examples
A first order IVP can have
1 NO solution : |y ′|+ |y | = 0, y(0) = 3.
2 Precisely one solution : y ′ = x , y(0) = 1. What is thesolution?
3 Infinitely many solutions: xy ′ = y − 1, y(0) = 1 The solutionsare y = 1 + cx .
Motivation to study conditions under which the solution wouldexist and the conditions under which it will be unique!
We first start with a few methods for finding out the solution offirst order ODEs, discuss the geometric meaning of solutions andthen proceed to study existence-uniqueness results.
Santanu Dey Lecture 2
Examples
A first order IVP can have
1 NO solution : |y ′|+ |y | = 0, y(0) = 3.
2 Precisely one solution : y ′ = x , y(0) = 1. What is thesolution?
3 Infinitely many solutions: xy ′ = y − 1, y(0) = 1 The solutionsare y = 1 + cx .
Motivation to study conditions under which the solution wouldexist and the conditions under which it will be unique!
We first start with a few methods for finding out the solution offirst order ODEs, discuss the geometric meaning of solutions andthen proceed to study existence-uniqueness results.
Santanu Dey Lecture 2
Examples
A first order IVP can have
1 NO solution : |y ′|+ |y | = 0, y(0) = 3.
2 Precisely one solution : y ′ = x , y(0) = 1. What is thesolution?
3 Infinitely many solutions: xy ′ = y − 1, y(0) = 1 The solutionsare y = 1 + cx .
Motivation to study conditions under which the solution wouldexist and the conditions under which it will be unique!
We first start with a few methods for finding out the solution offirst order ODEs, discuss the geometric meaning of solutions andthen proceed to study existence-uniqueness results.
Santanu Dey Lecture 2
Separable ODE’s
An ODE of the form
M(x) + N(y)y ′ = 0
is called a separable ODE.
Let H1(x) and H2(y) be any functions such that H ′1(x) = M(x)and H ′2(y) = N(y).Substituting in the DE, we obtain
H ′1(x) + H ′2(y)y ′ = 0.
Using chain rule,d
dxH2(y) = H ′2(y)
dy
dx.
Hence,d
dx(H1(x) + H2(y)) = 0.
Integrating, H1(x) + H2(y) = c, where c is an arbirtaryconstant.Note: This method many times gives us an implicit solution andnot necessarily an explicit one!
Santanu Dey Lecture 2
Separable ODE’s
An ODE of the form
M(x) + N(y)y ′ = 0
is called a separable ODE.Let H1(x) and H2(y) be any functions such that H ′1(x) = M(x)and H ′2(y) = N(y).
Substituting in the DE, we obtain
H ′1(x) + H ′2(y)y ′ = 0.
Using chain rule,d
dxH2(y) = H ′2(y)
dy
dx.
Hence,d
dx(H1(x) + H2(y)) = 0.
Integrating, H1(x) + H2(y) = c, where c is an arbirtaryconstant.Note: This method many times gives us an implicit solution andnot necessarily an explicit one!
Santanu Dey Lecture 2
Separable ODE’s
An ODE of the form
M(x) + N(y)y ′ = 0
is called a separable ODE.Let H1(x) and H2(y) be any functions such that H ′1(x) = M(x)and H ′2(y) = N(y).Substituting in the DE, we obtain
H ′1(x) + H ′2(y)y ′ = 0.
Using chain rule,d
dxH2(y) = H ′2(y)
dy
dx.
Hence,d
dx(H1(x) + H2(y)) = 0.
Integrating, H1(x) + H2(y) = c, where c is an arbirtaryconstant.Note: This method many times gives us an implicit solution andnot necessarily an explicit one!
Santanu Dey Lecture 2
Separable ODE’s
An ODE of the form
M(x) + N(y)y ′ = 0
is called a separable ODE.Let H1(x) and H2(y) be any functions such that H ′1(x) = M(x)and H ′2(y) = N(y).Substituting in the DE, we obtain
H ′1(x) + H ′2(y)y ′ = 0.
Using chain rule,d
dxH2(y) = H ′2(y)
dy
dx.
Hence,d
dx(H1(x) + H2(y)) = 0.
Integrating, H1(x) + H2(y) = c, where c is an arbirtaryconstant.Note: This method many times gives us an implicit solution andnot necessarily an explicit one!
Santanu Dey Lecture 2
Separable ODE’s
An ODE of the form
M(x) + N(y)y ′ = 0
is called a separable ODE.Let H1(x) and H2(y) be any functions such that H ′1(x) = M(x)and H ′2(y) = N(y).Substituting in the DE, we obtain
H ′1(x) + H ′2(y)y ′ = 0.
Using chain rule,d
dxH2(y) = H ′2(y)
dy
dx.
Hence,d
dx(H1(x) + H2(y)) = 0.
Integrating, H1(x) + H2(y) = c, where c is an arbirtaryconstant.Note: This method many times gives us an implicit solution andnot necessarily an explicit one!
Santanu Dey Lecture 2
Separable ODE’s
An ODE of the form
M(x) + N(y)y ′ = 0
is called a separable ODE.Let H1(x) and H2(y) be any functions such that H ′1(x) = M(x)and H ′2(y) = N(y).Substituting in the DE, we obtain
H ′1(x) + H ′2(y)y ′ = 0.
Using chain rule,d
dxH2(y) = H ′2(y)
dy
dx.
Hence,d
dx(H1(x) + H2(y)) = 0.
Integrating, H1(x) + H2(y) = c, where c is an arbirtaryconstant.Note:
This method many times gives us an implicit solution andnot necessarily an explicit one!
Santanu Dey Lecture 2
Separable ODE’s
An ODE of the form
M(x) + N(y)y ′ = 0
is called a separable ODE.Let H1(x) and H2(y) be any functions such that H ′1(x) = M(x)and H ′2(y) = N(y).Substituting in the DE, we obtain
H ′1(x) + H ′2(y)y ′ = 0.
Using chain rule,d
dxH2(y) = H ′2(y)
dy
dx.
Hence,d
dx(H1(x) + H2(y)) = 0.
Integrating, H1(x) + H2(y) = c, where c is an arbirtaryconstant.Note: This method many times gives us an implicit solution andnot necessarily an explicit one!
Santanu Dey Lecture 2
Separable ODE - Example 1
Solve the DE :y ′ = −2xy .
Separating the variables, we get :
dy
y= −2xdx .
Integrating both sides, we obtain :
ln |y | = −x2 + c1.
Thus, the solutions arey = ce−x
2.
How do they look?
Santanu Dey Lecture 2
Separable ODE - Example 1
Solve the DE :y ′ = −2xy .
Separating the variables, we get :
dy
y= −2xdx .
Integrating both sides, we obtain :
ln |y | = −x2 + c1.
Thus, the solutions arey = ce−x
2.
How do they look?
Santanu Dey Lecture 2
Separable ODE - Example 1
Solve the DE :y ′ = −2xy .
Separating the variables, we get :
dy
y= −2xdx .
Integrating both sides, we obtain :
ln |y | = −x2 + c1.
Thus, the solutions arey = ce−x
2.
How do they look?
Santanu Dey Lecture 2
Separable ODE - Example 1
Solve the DE :y ′ = −2xy .
Separating the variables, we get :
dy
y= −2xdx .
Integrating both sides, we obtain :
ln |y | = −x2 + c1.
Thus, the solutions arey = ce−x
2.
How do they look?
Santanu Dey Lecture 2
Separable ODE - Example 1
Solve the DE :y ′ = −2xy .
Separating the variables, we get :
dy
y= −2xdx .
Integrating both sides, we obtain :
ln |y | = −x2 + c1.
Thus, the solutions arey = ce−x
2.
How do they look?
Santanu Dey Lecture 2
Separable ODE - Example 1
Solve the DE :y ′ = −2xy .
Separating the variables, we get :
dy
y= −2xdx .
Integrating both sides, we obtain :
ln |y | = −x2 + c1.
Thus, the solutions are
y = ce−x2.
How do they look?
Santanu Dey Lecture 2
Separable ODE - Example 1
Solve the DE :y ′ = −2xy .
Separating the variables, we get :
dy
y= −2xdx .
Integrating both sides, we obtain :
ln |y | = −x2 + c1.
Thus, the solutions arey = ce−x
2.
How do they look?
Santanu Dey Lecture 2
Separable ODE - Example 1
Solve the DE :y ′ = −2xy .
Separating the variables, we get :
dy
y= −2xdx .
Integrating both sides, we obtain :
ln |y | = −x2 + c1.
Thus, the solutions arey = ce−x
2.
How do they look?
Santanu Dey Lecture 2
If we are given an initial condition
y(x0) = y0,
then we get:c = y0e
x20
and y = y0ex20−x2 .
(y0 = e−x20 )
Santanu Dey Lecture 2
If we are given an initial condition
y(x0) = y0,
then we get:c = y0e
x20
and y = y0ex20−x2 .
(y0 = e−x20 )
Santanu Dey Lecture 2
If we are given an initial condition
y(x0) = y0,
then we get:c = y0e
x20
and y = y0ex20−x2 .
(y0 = e−x20 )
Santanu Dey Lecture 2
Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx=
y cos x
1 + 2y2; y(0) = 1.
Assume y 6= 0. Then,
1 + 2y2
ydy = cos x dx .
Integrating,ln |y |+ y2 = sin x + c .
As y(0) = 1, we get c = 1. Hence a particular solution to the IVPis
ln |y |+ y2 = sin x + 1.
Note: y ≡ 0 is a solution to the DE but it is not a solution to thegiven IVP.
Santanu Dey Lecture 2
Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx=
y cos x
1 + 2y2; y(0) = 1.
Assume y 6= 0.
Then,
1 + 2y2
ydy = cos x dx .
Integrating,ln |y |+ y2 = sin x + c .
As y(0) = 1, we get c = 1. Hence a particular solution to the IVPis
ln |y |+ y2 = sin x + 1.
Note: y ≡ 0 is a solution to the DE but it is not a solution to thegiven IVP.
Santanu Dey Lecture 2
Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx=
y cos x
1 + 2y2; y(0) = 1.
Assume y 6= 0. Then,
1 + 2y2
ydy = cos x dx .
Integrating,ln |y |+ y2 = sin x + c .
As y(0) = 1, we get c = 1. Hence a particular solution to the IVPis
ln |y |+ y2 = sin x + 1.
Note: y ≡ 0 is a solution to the DE but it is not a solution to thegiven IVP.
Santanu Dey Lecture 2
Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx=
y cos x
1 + 2y2; y(0) = 1.
Assume y 6= 0. Then,
1 + 2y2
ydy = cos x dx .
Integrating,
ln |y |+ y2 = sin x + c .
As y(0) = 1, we get c = 1. Hence a particular solution to the IVPis
ln |y |+ y2 = sin x + 1.
Note: y ≡ 0 is a solution to the DE but it is not a solution to thegiven IVP.
Santanu Dey Lecture 2
Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx=
y cos x
1 + 2y2; y(0) = 1.
Assume y 6= 0. Then,
1 + 2y2
ydy = cos x dx .
Integrating,ln |y |+ y2 = sin x + c .
As y(0) = 1, we get c = 1. Hence a particular solution to the IVPis
ln |y |+ y2 = sin x + 1.
Note: y ≡ 0 is a solution to the DE but it is not a solution to thegiven IVP.
Santanu Dey Lecture 2
Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx=
y cos x
1 + 2y2; y(0) = 1.
Assume y 6= 0. Then,
1 + 2y2
ydy = cos x dx .
Integrating,ln |y |+ y2 = sin x + c .
As y(0) = 1, we get c = 1.
Hence a particular solution to the IVPis
ln |y |+ y2 = sin x + 1.
Note: y ≡ 0 is a solution to the DE but it is not a solution to thegiven IVP.
Santanu Dey Lecture 2
Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx=
y cos x
1 + 2y2; y(0) = 1.
Assume y 6= 0. Then,
1 + 2y2
ydy = cos x dx .
Integrating,ln |y |+ y2 = sin x + c .
As y(0) = 1, we get c = 1. Hence a particular solution to the IVPis
ln |y |+ y2 = sin x + 1.
Note: y ≡ 0 is a solution to the DE but it is not a solution to thegiven IVP.
Santanu Dey Lecture 2
Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx=
y cos x
1 + 2y2; y(0) = 1.
Assume y 6= 0. Then,
1 + 2y2
ydy = cos x dx .
Integrating,ln |y |+ y2 = sin x + c .
As y(0) = 1, we get c = 1. Hence a particular solution to the IVPis
ln |y |+ y2 = sin x + 1.
Note: y ≡ 0 is a solution to the DE
but it is not a solution to thegiven IVP.
Santanu Dey Lecture 2
Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx=
y cos x
1 + 2y2; y(0) = 1.
Assume y 6= 0. Then,
1 + 2y2
ydy = cos x dx .
Integrating,ln |y |+ y2 = sin x + c .
As y(0) = 1, we get c = 1. Hence a particular solution to the IVPis
ln |y |+ y2 = sin x + 1.
Note: y ≡ 0 is a solution to the DE but it is not a solution to thegiven IVP.
Santanu Dey Lecture 2
Method of separation of variables doesn’t yield allsolutions!
Solve y ′ = 3y2/3, y(0) = 0.
y ≡ 0 is a solution.
If y 6= 0,dy
y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.
Initial condition yields c = 0.
Hence y = x3 and y = 0 are solutions which satisfy the initialconditions.Consider
φk(x) =
{0 −∞ < x ≤ k
(x − k)3 k < x <∞
Are these functions solutions of the DE? YES.There are infinitely many functions which are solutions of the DE.
Santanu Dey Lecture 2
Method of separation of variables doesn’t yield allsolutions!
Solve y ′ = 3y2/3, y(0) = 0.y ≡ 0 is a solution.
If y 6= 0,dy
y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.
Initial condition yields c = 0.
Hence y = x3 and y = 0 are solutions which satisfy the initialconditions.Consider
φk(x) =
{0 −∞ < x ≤ k
(x − k)3 k < x <∞
Are these functions solutions of the DE? YES.There are infinitely many functions which are solutions of the DE.
Santanu Dey Lecture 2
Method of separation of variables doesn’t yield allsolutions!
Solve y ′ = 3y2/3, y(0) = 0.y ≡ 0 is a solution.
If y 6= 0,dy
y2/3= 3dx
=⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.
Initial condition yields c = 0.
Hence y = x3 and y = 0 are solutions which satisfy the initialconditions.Consider
φk(x) =
{0 −∞ < x ≤ k
(x − k)3 k < x <∞
Are these functions solutions of the DE? YES.There are infinitely many functions which are solutions of the DE.
Santanu Dey Lecture 2
Method of separation of variables doesn’t yield allsolutions!
Solve y ′ = 3y2/3, y(0) = 0.y ≡ 0 is a solution.
If y 6= 0,dy
y2/3= 3dx =⇒ 3y1/3 = 3(x + c)
=⇒ y = (x + c)3.
Initial condition yields c = 0.
Hence y = x3 and y = 0 are solutions which satisfy the initialconditions.Consider
φk(x) =
{0 −∞ < x ≤ k
(x − k)3 k < x <∞
Are these functions solutions of the DE? YES.There are infinitely many functions which are solutions of the DE.
Santanu Dey Lecture 2
Method of separation of variables doesn’t yield allsolutions!
Solve y ′ = 3y2/3, y(0) = 0.y ≡ 0 is a solution.
If y 6= 0,dy
y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.
Initial condition yields c = 0.
Hence y = x3 and y = 0 are solutions which satisfy the initialconditions.Consider
φk(x) =
{0 −∞ < x ≤ k
(x − k)3 k < x <∞
Are these functions solutions of the DE? YES.There are infinitely many functions which are solutions of the DE.
Santanu Dey Lecture 2
Method of separation of variables doesn’t yield allsolutions!
Solve y ′ = 3y2/3, y(0) = 0.y ≡ 0 is a solution.
If y 6= 0,dy
y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.
Initial condition yields c = 0.
Hence y = x3 and y = 0 are solutions which satisfy the initialconditions.Consider
φk(x) =
{0 −∞ < x ≤ k
(x − k)3 k < x <∞
Are these functions solutions of the DE? YES.There are infinitely many functions which are solutions of the DE.
Santanu Dey Lecture 2
Method of separation of variables doesn’t yield allsolutions!
Solve y ′ = 3y2/3, y(0) = 0.y ≡ 0 is a solution.
If y 6= 0,dy
y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.
Initial condition yields c = 0.
Hence y = x3 and y = 0 are solutions which satisfy the initialconditions.
Consider
φk(x) =
{0 −∞ < x ≤ k
(x − k)3 k < x <∞
Are these functions solutions of the DE? YES.There are infinitely many functions which are solutions of the DE.
Santanu Dey Lecture 2
Method of separation of variables doesn’t yield allsolutions!
Solve y ′ = 3y2/3, y(0) = 0.y ≡ 0 is a solution.
If y 6= 0,dy
y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.
Initial condition yields c = 0.
Hence y = x3 and y = 0 are solutions which satisfy the initialconditions.Consider
φk(x) =
{0 −∞ < x ≤ k
(x − k)3 k < x <∞
Are these functions solutions of the DE? YES.There are infinitely many functions which are solutions of the DE.
Santanu Dey Lecture 2
Method of separation of variables doesn’t yield allsolutions!
Solve y ′ = 3y2/3, y(0) = 0.y ≡ 0 is a solution.
If y 6= 0,dy
y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.
Initial condition yields c = 0.
Hence y = x3 and y = 0 are solutions which satisfy the initialconditions.Consider
φk(x) =
{0 −∞ < x ≤ k
(x − k)3 k < x <∞
Are these functions solutions of the DE?
YES.There are infinitely many functions which are solutions of the DE.
Santanu Dey Lecture 2
Method of separation of variables doesn’t yield allsolutions!
Solve y ′ = 3y2/3, y(0) = 0.y ≡ 0 is a solution.
If y 6= 0,dy
y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.
Initial condition yields c = 0.
Hence y = x3 and y = 0 are solutions which satisfy the initialconditions.Consider
φk(x) =
{0 −∞ < x ≤ k
(x − k)3 k < x <∞
Are these functions solutions of the DE? YES.There are infinitely many functions which are solutions of the DE.
Santanu Dey Lecture 2
Method of separation of variables doesn’t yield allsolutions!
Solve y ′ = 3y2/3, y(0) = 0.y ≡ 0 is a solution.
If y 6= 0,dy
y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.
Initial condition yields c = 0.
Hence y = x3 and y = 0 are solutions which satisfy the initialconditions.Consider
φk(x) =
{0 −∞ < x ≤ k
(x − k)3 k < x <∞
Are these functions solutions of the DE? YES.There are infinitely many functions which are solutions of the DE.
Santanu Dey Lecture 2
Homogeneous functions
Definition
A function f (x1, . . . , xn) is called homogeneous if
f (tx1, . . . , txn) = td f (x1, . . . , xn)
for some d ∈ Z and for all t 6= 0.
The number d is called the degree of f (x1, . . . , xn).Examples :f (x , y) = x2 + xy + y2 is homogeneous of degree 2.
f (x , y) = y + x cos2(yx
)is homogeneous of degree 1.
Santanu Dey Lecture 2
Homogeneous functions
Definition
A function f (x1, . . . , xn) is called homogeneous if
f (tx1, . . . , txn) = td f (x1, . . . , xn)
for some d ∈ Z and for all t 6= 0.
The number d is called the degree of f (x1, . . . , xn).
Examples :f (x , y) = x2 + xy + y2 is homogeneous of degree 2.
f (x , y) = y + x cos2(yx
)is homogeneous of degree 1.
Santanu Dey Lecture 2
Homogeneous functions
Definition
A function f (x1, . . . , xn) is called homogeneous if
f (tx1, . . . , txn) = td f (x1, . . . , xn)
for some d ∈ Z and for all t 6= 0.
The number d is called the degree of f (x1, . . . , xn).Examples :f (x , y) = x2 + xy + y2 is homogeneous of degree
2.
f (x , y) = y + x cos2(yx
)is homogeneous of degree 1.
Santanu Dey Lecture 2
Homogeneous functions
Definition
A function f (x1, . . . , xn) is called homogeneous if
f (tx1, . . . , txn) = td f (x1, . . . , xn)
for some d ∈ Z and for all t 6= 0.
The number d is called the degree of f (x1, . . . , xn).Examples :f (x , y) = x2 + xy + y2 is homogeneous of degree 2.
f (x , y) = y + x cos2(yx
)is homogeneous of degree 1.
Santanu Dey Lecture 2
Homogeneous functions
Definition
A function f (x1, . . . , xn) is called homogeneous if
f (tx1, . . . , txn) = td f (x1, . . . , xn)
for some d ∈ Z and for all t 6= 0.
The number d is called the degree of f (x1, . . . , xn).Examples :f (x , y) = x2 + xy + y2 is homogeneous of degree 2.
f (x , y) = y + x cos2(yx
)is homogeneous of degree
1.
Santanu Dey Lecture 2
Homogeneous functions
Definition
A function f (x1, . . . , xn) is called homogeneous if
f (tx1, . . . , txn) = td f (x1, . . . , xn)
for some d ∈ Z and for all t 6= 0.
The number d is called the degree of f (x1, . . . , xn).Examples :f (x , y) = x2 + xy + y2 is homogeneous of degree 2.
f (x , y) = y + x cos2(yx
)is homogeneous of degree 1.
Santanu Dey Lecture 2
Homogeneous functions
Definition
A function f (x1, . . . , xn) is called homogeneous if
f (tx1, . . . , txn) = td f (x1, . . . , xn)
for some d ∈ Z and for all t 6= 0.
The number d is called the degree of f (x1, . . . , xn).Examples :f (x , y) = x2 + xy + y2 is homogeneous of degree 2.
f (x , y) = y + x cos2(yx
)is homogeneous of degree 1.
Santanu Dey Lecture 2
Homogeneous Equations
Definition
The first order ODE
M(x , y) + N(x , y)dy
dx= 0
is called homogeneous if M and N are homogeneous of equaldegree.
Example :
(y2 − x2)dy
dx+ 2xy = 0.
Santanu Dey Lecture 2
Homogeneous ODE’s - Reduction to variable separableform
Let
M(x , y) + N(x , y)dy
dx= 0
where M and N are homogeneous of degree d .
Put
y
x= v .
Then,dy
dx= x
dv
dx+ v .
Substituting this in the given ODE, we get:
M(x , xv) + N(x , xv)
(xdv
dx+ v
)= 0.
Thus,
xdM(1, v) + xdN(1, v)
(xdv
dx+ v
)= 0.
Santanu Dey Lecture 2
Homogeneous ODE’s - Reduction to variable separableform
Let
M(x , y) + N(x , y)dy
dx= 0
where M and N are homogeneous of degree d . Put
y
x= v .
Then,dy
dx= x
dv
dx+ v .
Substituting this in the given ODE, we get:
M(x , xv) + N(x , xv)
(xdv
dx+ v
)= 0.
Thus,
xdM(1, v) + xdN(1, v)
(xdv
dx+ v
)= 0.
Santanu Dey Lecture 2
Homogeneous ODE’s - Reduction to variable separableform
Let
M(x , y) + N(x , y)dy
dx= 0
where M and N are homogeneous of degree d . Put
y
x= v .
Then,dy
dx= x
dv
dx+ v .
Substituting this in the given ODE, we get:
M(x , xv) + N(x , xv)
(xdv
dx+ v
)= 0.
Thus,
xdM(1, v) + xdN(1, v)
(xdv
dx+ v
)= 0.
Santanu Dey Lecture 2
Homogeneous ODE’s - Reduction to variable separableform
Let
M(x , y) + N(x , y)dy
dx= 0
where M and N are homogeneous of degree d . Put
y
x= v .
Then,dy
dx= x
dv
dx+ v .
Substituting this in the given ODE, we get:
M(x , xv) + N(x , xv)
(xdv
dx+ v
)= 0.
Thus,
xdM(1, v) + xdN(1, v)
(xdv
dx+ v
)= 0.
Santanu Dey Lecture 2
Homogeneous ODE’s - Reduction to variable separableform
Let
M(x , y) + N(x , y)dy
dx= 0
where M and N are homogeneous of degree d . Put
y
x= v .
Then,dy
dx= x
dv
dx+ v .
Substituting this in the given ODE, we get:
M(x , xv) + N(x , xv)
(xdv
dx+ v
)= 0.
Thus,
xdM(1, v) + xdN(1, v)
(xdv
dx+ v
)= 0.
Santanu Dey Lecture 2
Homogeneous ODE’s - Reduction to variable separableform
Let
M(x , y) + N(x , y)dy
dx= 0
where M and N are homogeneous of degree d . Put
y
x= v .
Then,dy
dx= x
dv
dx+ v .
Substituting this in the given ODE, we get:
M(x , xv) + N(x , xv)
(xdv
dx+ v
)= 0.
Thus,
xdM(1, v) + xdN(1, v)
(xdv
dx+ v
)= 0.
Santanu Dey Lecture 2
Homogeneous ODE’s
Let x 6= 0.
Then,
M(1, v) + N(1, v) · v + N(1, v) · x dvdx
= 0.
Thus,dx
x+
N(1, v)
M(1, v) + N(1, v) · vdv = 0.
This is a separable equation.
Santanu Dey Lecture 2
Homogeneous ODE’s
Let x 6= 0. Then,
M(1, v) + N(1, v) · v + N(1, v) · x dvdx
= 0.
Thus,dx
x+
N(1, v)
M(1, v) + N(1, v) · vdv = 0.
This is a separable equation.
Santanu Dey Lecture 2
Homogeneous ODE’s
Let x 6= 0. Then,
M(1, v) + N(1, v) · v + N(1, v) · x dvdx
= 0.
Thus,dx
x+
N(1, v)
M(1, v) + N(1, v) · vdv = 0.
This is a separable equation.
Santanu Dey Lecture 2
Homogeneous ODE’s
Let x 6= 0. Then,
M(1, v) + N(1, v) · v + N(1, v) · x dvdx
= 0.
Thus,dx
x+
N(1, v)
M(1, v) + N(1, v) · vdv = 0.
This is a separable equation.
Santanu Dey Lecture 2