y′−y=exy2
y2y′−y1=ex
Use the change of variable
z=y1−2=y−1
z=y−1 Differentiate both sides with respect to x
z′=−y−2y′ Then
y′=−z′y2=−z′z−2 Substitute
z′+z=−exWe get the linear equation for the function z(x). To solve it, we use the integrating factor:
u(x)=e∫(1)dx=ex
exz′+exz=−e2x
d(zex)=−e2xdx Integrate
∫d(zex)=−∫e2xdx
zex=−21e2x+21C
z=−21ex+21Ce−x Then
y=−21ex+21Ce−x1,y=0
y=−ex+Ce−x2,or y=0
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