How to solve y′+y=exy23?

Karen Simpson

Karen Simpson

Answered

2022-01-19

How to solve y+y=exy23?

Answer & Explanation

Beverly Smith

Beverly Smith

Expert

2022-01-19Added 42 answers

There is one obvious solution y(x)=0.
y+y=exy23
yex+yex=e2xy23
(yex)=e2xy23
Substitute u(x)=y(x)ex
u=u23e4x3
u23u=e4x3
We have divided by u23, so this is ok only for points such that u(x)0.
13u23u=13e4x3
(u13)=(4x34)
u13=e4x34+C
u=(C+e4x34)3
y=(C+e4x34)3ex
To get y=0, we need C=14
y=(e4x314)3ex
Mary Nicholson

Mary Nicholson

Expert

2022-01-20Added 38 answers

The idea is that in differential equations of the form y+g(x)y=f(x)yα(α1), the Bernoulli equations, you can do the variable change y=uβ, and your equation becomes
βuuβ1+g(x)uβ=f(x)uβα
Multiplying through by 1βu1β, this is the same as
u+1βg(x)u=1βf(x)uβα+1β
So the idea is that if you choose β so that βα+1β=0, in other words, β=11α, your equation becomes first-order linear and you can solve it using first-order linear methods. In this case, α=23,so β=3 and you make the variable change y=u3

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