pot the Mistake! Using FOBDE find the general solution of: dydx−y=xy2dydx−y=xy2 Solution: P(x)=−1;Q(x)=x;n=2P(x)=−1;Q(x)=x;n=2 μ(x)=e(1−2)∫−1dxμ(x)=e(1−2)∫−1dx μ(x)=exμ(x)=ex y(1−2)ex=(1−2)∫xexdx+Cy(1−2)ex=(1−2)∫xexdx+C using integration by parts uv−∫vduuv−∫vdu ∫xexdx∫xexdx letu=xdv=exdxletu=xdv=exdx dv=1dxv=exdv=1dxv=ex xex−∫exdxxex−∫exdx xex−exxex−ex y(1−2)ex=(1−2)[xex−ex]+Cy(1−2)ex=(1−2)[xex−ex]+C y−1ex=ex(−x+1)+Cy−1ex=ex(−x+1)+C divide it both side by exex y−1=−x+1+Ce−x
pot the Mistake! Using FOBDE find the general solution of: dydx−y=xy2dydx−y=xy2 Solution: P(x)=−1;Q(x)=x;n=2P(x)=−1;Q(x)=x;n=2 μ(x)=e(1−2)∫−1dxμ(x)=e(1−2)∫−1dx μ(x)=exμ(x)=ex y(1−2)ex=(1−2)∫xexdx+Cy(1−2)ex=(1−2)∫xexdx+C using integration by parts uv−∫vduuv−∫vdu ∫xexdx∫xexdx letu=xdv=exdxletu=xdv=exdx dv=1dxv=exdv=1dxv=ex xex−∫exdxxex−∫exdx xex−exxex−ex y(1−2)ex=(1−2)[xex−ex]+Cy(1−2)ex=(1−2)[xex−ex]+C y−1ex=ex(−x+1)+Cy−1ex=ex(−x+1)+C divide it both side by exex y−1=−x+1+Ce−x
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Spot the Mistake!
Using FOBDE find the general solution of:
dydx−y=xy2dydx−y=xy2
Solution:
P(x)=−1;Q(x)=x;n=2P(x)=−1;Q(x)=x;n=2
μ(x)=e(1−2)∫−1dxμ(x)=e(1−2)∫−1dx
μ(x)=exμ(x)=ex
y(1−2)ex=(1−2)∫xexdx+Cy(1−2)ex=(1−2)∫xexdx+C
using
uv−∫vduuv−∫vdu
∫xexdx∫xexdx
letu=xdv=exdxletu=xdv=exdx
dv=1dxv=exdv=1dxv=ex
xex−∫exdxxex−∫exdx
xex−exxex−ex
y(1−2)ex=(1−2)[xex−ex]+Cy(1−2)ex=(1−2)[xex−ex]+C
y−1ex=ex(−x+1)+Cy−1ex=ex(−x+1)+C
divide it both side by exex
y−1=−x+1+Ce−x
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