Karen Simpson

Answered

2022-01-19

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

Answer & Explanation

Beverly Smith

Expert

2022-01-19Added 42 answers

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 $u\left(x\right)\ne 0$ .

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

$\left({u}^{\frac{1}{3}}\right)}^{\prime}={\left(\frac{\frac{4x}{3}}{4}\right)}^{\prime$

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

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

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

To get$y=0$ , we need $C=-\frac{1}{4}$

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

Substitute

We have divided by

To get

Mary Nicholson

Expert

2022-01-20Added 38 answers

The idea is that in differential equations of the form ${y}^{\prime}+g\left(x\right)y=f\left(x\right){y}^{\alpha}(\alpha \ne 1)$ , 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, $\alpha =\frac{2}{3},\text{so}\text{}\beta =3$ and you make the variable change $y={u}^{3}$

Multiplying through by

So the idea is that if you choose

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