Check the solution to "How to solve y′+y=exy23?"

Name: How to solve y′+y=exy23?
Uploaded: 2022-02-23T17:22:57+00:00
Description: How to solve y′+y=exy23?

Karen Simpson

Answered question

2022-01-19

How to solve

y^{'} + y = e^{x} y^{\frac{2}{3}}

Beverly Smith

Beginner2022-01-19Added 42 answers

There is one obvious solution

y (x) = 0

y^{'} + y = e^{x} y^{\frac{2}{3}}

y^{'} e^{x} + y e^{x} = e^{2 x} y^{\frac{2}{3}}

{(y e^{x})}^{'} = e^{2 x} y^{\frac{2}{3}}

Substitute

u (x) = y (x) e^{x}

u^{'} = u^{\frac{2}{3}} e^{\frac{4 x}{3}}

u^{- \frac{2}{3}} u^{'} = e^{\frac{4 x}{3}}

We have divided by

u^{\frac{2}{3}}

, so this is ok only for points such that

u (x) \neq 0

\frac{1}{3} u^{\frac{- 2}{3}} u^{'} = \frac{1}{3} e^{\frac{4 x}{3}}

{(u^{\frac{1}{3}})}^{'} = {(\frac{\frac{4 x}{3}}{4})}^{'}

u^{\frac{1}{3}} = \frac{e^{\frac{4 x}{3}}}{4} + C

u = {(C + \frac{e^{\frac{4 x}{3}}}{4})}^{3}

y = {(C + \frac{e^{\frac{4 x}{3}}}{4})}^{3} e^{- x}

To get

y = 0

, we need

C = - \frac{1}{4}

y = {(\frac{e^{4 \frac{x}{3}} - 1}{4})}^{3} e^{- x}

Mary Nicholson

Beginner2022-01-20Added 38 answers

The idea is that in differential equations of the form

y^{'} + g (x) y = f (x) y^{α} (α \neq 1)

, the Bernoulli equations, you can do the variable change

y = u^{β}

, and your equation becomes

β u^{'} u^{β - 1} + g (x) u^{β} = f (x) u^{β α}

Multiplying through by

\frac{1}{β} u^{1 - β}

, this is the same as

u^{'} + \frac{1}{β} g (x) u = \frac{1}{β} f (x) u^{β α + 1 - β}

So the idea is that if you choose

β

so that

β α + 1 - β = 0

, in other words,

β = \frac{1}{1 - α}

, your equation becomes first-order linear and you can solve it using first-order linear methods. In this case,

α = \frac{2}{3}, so β = 3

and you make the variable change

y = u^{3}

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