sec 1.5: linear first-order de
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Sec 1.5: Linear First-Order DE
1
Sec 1.5: Linear First-Order DE
Definition 2.2
:Example
)()(')( 01 xgyxayxa
1
A 1st order De of the form
Sec 2.3
is said to be linear first-order equation.
03' yy 2 363' yy xexyxy 64'
)()(' xQyxPy
2
How to Solve ?
Q(x)P(x)yy '
Sec 1.5
Method of Solution:
Step 1Rewrite into standard form
(coeff of y’ is 1).
dxxP
ex)(
)(Step 2Find the Integrating Factor =
( ignore constant of integration)
----- (1)
Step 3 Multiply (1) by ( check: )][ LHS yDy](x)[dx
dx
Step 4Integrate both sides:
)()( xQxy](x)[dx
d ( DONOT forget constant of integration)
:Example
1 03' yy 2 363' yy xexyxy 64'
x
x
3
Remark: linear in y or linear in x
:Example
Sec 2.3
1)0(8
11' 3/
y
eyy x
xxyyx 63')1( 2
Solve the IVP
:Example Find a general solution of
4
:Example Find a general solution of
2
1
yxdx
dy
Q(x)P(x)yy '
)()( )(
1)( CdxxxQ
xxy
The solution of the above DE is given:
dxxP
ex)(
)(where
Derivation
5
00 )(
'
yxy
Q(x)P(x)yy
)()( )(
1)( CdxxxQ
xxy
If the function and are continuous on the open interval I containing the point , then the initial value problem
dxxP
ex)(
)(where
Theorem 1:
Theorem 1: The Linear First-Order Equation
has a unique solution y(x) on I, given by the formula in Eq(6) with an appropriate value of C.
xP xQ
0x
(6)
Remarks:
1. Theorem 1 gives a solution on the entire interval I
2. Theorem 1 tells us every solution is included in the formula (6)
3. Theorem 1 tells us that the general solution is given in(6)
4. Theorem 1 tells us that a linear first-order DE has no singular sol
6
:Example
Sec 2.3
10
2
)1(
sin'
yy
xxyyx
Solve the IVP
7
Derivative and Integration
D[x^3+3 x,{x,2}]
x][xdx
d33
2
2
Integrate[ x^2 , x]
dxx2
syms x
diff(x^3+3 x)
MA
TH
EM
AT
ICA
syms x
int(x^2,x)
MA
TL
AB
8
9
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