A)Givenϕ (x) = xe2x
ϕ′=e2x(1+2x)
If ϕ(x) is a solution to the given equation then it satisfies the equation,
e2x(1+2x)+axe2x−e2x=01+2x+ax−1=02x+ax=0a=−2
B)
Given y(ex+1)=x-1
Differentiating with respect to x
dy ( ex+1) + y(ex) dx = dx
dy(ex+1) = dx ( 1-yex)
dxdy=ex+11−yex
Putting this value in the given differential equation
ex+11−yex=e−xy+e−xyay+x
ex+11−yex−e−xy−x =e−xa
1+ex1−ye−x−y(1+ex)−x(1+ex)=aexex(1+ex)1−ye−x−(1+ex)(x+y)=a
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