Karen Simpson

2022-01-19

How to solve ${y}^{\prime }+y={e}^{x}{y}^{\frac{2}{3}}$?

Beverly Smith

Expert

There is one obvious solution $y\left(x\right)=0$.
${y}^{\prime }+y={e}^{x}{y}^{\frac{2}{3}}$
${y}^{\prime }{e}^{x}+y{e}^{x}={e}^{2x}{y}^{\frac{2}{3}}$
${\left(y{e}^{x}\right)}^{\prime }={e}^{2x}{y}^{\frac{2}{3}}$
Substitute $u\left(x\right)=y\left(x\right){e}^{x}$
${u}^{\prime }={u}^{\frac{2}{3}}{e}^{\frac{4x}{3}}$
${u}^{-\frac{2}{3}}{u}^{\prime }={e}^{\frac{4x}{3}}$
We have divided by ${u}^{\frac{2}{3}}$, so this is ok only for points such that $u\left(x\right)\ne 0$.
$\frac{1}{3}{u}^{\frac{-2}{3}}{u}^{\prime }=\frac{1}{3}{e}^{\frac{4x}{3}}$
${\left({u}^{\frac{1}{3}}\right)}^{\prime }={\left(\frac{\frac{4x}{3}}{4}\right)}^{\prime }$
${u}^{\frac{1}{3}}=\frac{{e}^{\frac{4x}{3}}}{4}+C$
$u={\left(C+\frac{{e}^{\frac{4x}{3}}}{4}\right)}^{3}$
$y={\left(C+\frac{{e}^{\frac{4x}{3}}}{4}\right)}^{3}{e}^{-x}$
To get $y=0$, we need $C=-\frac{1}{4}$
$y={\left(\frac{{e}^{4\frac{x}{3}}-1}{4}\right)}^{3}{e}^{-x}$

Mary Nicholson

Expert

The idea is that in differential equations of the form ${y}^{\prime }+g\left(x\right)y=f\left(x\right){y}^{\alpha }\left(\alpha \ne 1\right)$, the Bernoulli equations, you can do the variable change $y={u}^{\beta }$, and your equation becomes
$\beta {u}^{\prime }{u}^{\beta -1}+g\left(x\right){u}^{\beta }=f\left(x\right){u}^{\beta \alpha }$
Multiplying through by $\frac{1}{\beta }{u}^{1-\beta }$, this is the same as
${u}^{\prime }+\frac{1}{\beta }g\left(x\right)u=\frac{1}{\beta }f\left(x\right){u}^{\beta \alpha +1-\beta }$
So the idea is that if you choose $\beta$ so that $\beta \alpha +1-\beta =0$, in other words, $\beta =\frac{1}{1-\alpha }$, your equation becomes first-order linear and you can solve it using first-order linear methods. In this case, and you make the variable change $y={u}^{3}$

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