Question #118562
Answer the following and give complete solutions.

1. Find the constant a so that phi(x)=x(e^2x) is a solution to y'+ay-e^(2x)=0
2. Find the constant a so that e^xy + y = x-1 is an implicit solution to dy/dx = e^-xy +ay / e^-xy + x.
1
Expert's answer
2020-05-31T19:01:54-0400

A)Givenϕ (x) = xe2x


ϕ=e2x(1+2x)\phi '= e^{2x}(1 + 2x)

If  ϕ(x) is a solution to the given equation then it satisfies the equation,

e2x(1+2x)+axe2xe2x=01+2x+ax1=02x+ax=0a=2e^{2x}(1+2x)+axe^{2x}-e^{2x}=0 \\ 1+2x+ax-1=0 \\ 2x+ax=0 \\ a=-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)

dydx=1yexex+1\frac{dy}{dx} = \frac{1-ye^{x}}{e^{x} +1}


Putting this value in the given differential equation

1yexex+1=exy+ayexy+x\frac{1-ye^{x}}{e^{x} +1}=e^{-x}y+\frac{ay}{e^{-x}y}+x\\


1yexex+1exyx\frac{1-ye^{x}}{e^{x} +1}-e^{-x}y-x =aex\frac{a}{e^{-x}}

 

1yexy(1+ex)x(1+ex)1+ex=aex1yex(1+ex)(x+y)ex(1+ex)=a\frac{1-ye^{-x}-y(1+e^x)-x(1+e^x)}{1+e^x}= ae^x \\ \\ \frac{1-ye^{-x}-(1+e^x)(x+y)}{e^x(1+e^x)}= a


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