First Order Differential Equations: General Solutions, Particular Solutions and Separable Equations

General Solutions: A general solution to an nth order differential equation is a solution in which the value of the constant, C in the solution, may vary. For example, the general solution to the differential equation

1−= xedx

dy

is

Cxey x +−= . For instance, y = ex-x-5 is a solution to the above differential equation because C = 5. By differentiating y, you will see that the derivative of a constant is zero. Other solutions to the above differential equation could be:

4+−= xey x 5+−= xey x 54.100+−= xey x Particular Solutions: A particular solution to an nth order differential equation is a solution in which there is a particular value for the constant C. These types of solutions occur in initial value problems. An example of an initial value problem is given below.

1−= xedx

dy 3)0( =y

We know that the general solution to the differential equation is

Cxey x +−= . To find the particular solution, use the initial condition given to solve for C.

C

C

C

Ce

=+=

+−=+−=

2

13

013

03 0

Particular Solution:

2+−= xey x

Separable Equations: A separable equation is a first order differential equation where the independent variable is on one side of the equation and the dependent variable is on the other side of the equation with their respective differentials. It can be written as the following:

dxxgdyyh )()( = where h(y) = 1/p(y), which is a function of y alone and g(x) is a function of x alone. Steps To Solving A Separable Equation: 1. Write the differential equation in the form

dxxgdyyh )()( = where h(y) = 1/p(y).

2. Integrate both sides of the equation.

CxGyH

dxxgdyyh

+=

= ∫∫)()(

)()(

3. If necessary, solve for y. Example One: Solve

3

1

xydx

dy = .

1. Separate the variables. --Multiply both sides by dx.

dxxy

dy3

1=

--Multiply both sides by y3.

dxx

dyy13 =

2. Integrate both sides.

Cxy

dxx

dyy

+=

= ∫∫ln

4

1

4

3

3. (Optional) Solve for y.

( ) 4/1

4

4

ln4

ln4

ln4

Cxy

Cxy

Cxy

+±=

+=

+=

Implicit General Solution: This answer is not completely solved for y, but it is still a correct solution.

Explicit General Solution: This answer is completely solved for y. It is still a correct solution.

Example Two: Solve the initial value problem

xydx

dytan)1( 2+= 3)0( =y .

1. Separate the variables. --Multiply both sides by dx.

xdxydy tan)1( 2+= --Multiply both sides by y3.

xdxdyy tan)1( 2 =+ 2. Integrate both sides.

Cxy

y

xdxdyydy

xdxdyy

+−=+

=+

=+

∫∫ ∫∫∫

cosln2

tan

tan)1(

2

2

2

3. (Optional) Solve for y. --In this solution, it will be fine to leave as an implicit function because it will be difficult to solve for y.

4. Use the initial condition to find a particular solution to the above differential equation.

--Substitute 0 in for x and 3 for y to solve for C.

( )

C

C

C

C

C

=+

+=+

+=+

+−=+

+−=+

2

332

02

33

02

33

1ln2

33

0cosln2

33

2

Therefore, the general solution is

2

332cosln

2

2 ++−=+ xy

y