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$$y'=y+\frac{sin2x}{y^2}$$

The equation is clearly Bernoulli Differential Equation:

$y'=y +\sin2x y^{-2}$

But to which form should I get the equation?

1.leave it and use $z=y^{-3}$

  1. multiply by $y^2$ do get $y^2y'=y^3+sin2x$

To be more precise I have done the following:

$$y'=y+\frac{sin2x}{y^2}$$

$$y'=y+sin2x\cdot y^{-2}$$

$$z=y^{-3}$$

$$z'=-3y^{-4}y'=-3y^{-4}(y+sin2x\cdot y^{-2})=-3y^{-3}-3sin2xy^{-6}=-3z-3sin2xz^2$$

So I ended up with $z'=-3z-3sin2x z^2$ which is not linear, while the answer which started with option 2 above ended up with a linear ODE

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    @Moo I added what I have done, why should I multiple by $3$? in general to multiply by the power $n$ of $y$ in the substitution $v=y^n$? – 2017-01-25
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    I multiplied by $3$, because I noticed that we have $y^2 y'$ and if I took the derivative of $y^3$, that would give $3 y^2 y'$. This gives us the form $v' - 3 v = 3 \sin 2x$, which is MUCH easier to work with, because I can solve that many ways, like Integrating Factor. **Note:** I had a slight typo in my first comment and meant $3y^2 y' - 3 y^3 = 3 \sin 2x$. Then we let $v = y^3$ and this reduces the equation to $v' - 3 v = 3 \sin 2x$. – 2017-01-25
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    You should made the wrong substitution it should be $z=y^{1--2}$ – 2017-01-25

1 Answers 1

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You need to make the substitution $z=y^3\to z^{1/3}=y\implies\frac{1}{3}z^{-2/3}z'=y'$ rewriting the differential equation as $\frac{1}{3}z^{-2/3}z'=z^{1/3}+z^{-2/3}\sin(2x)$ multiplying through by $z^{2/3}$ we get $\frac{1}{3}z'=z+\sin(2x)$ can you proceed from here?

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    Shouldn't it be easier to to multiply by 3? – 2017-01-25
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    @gbox yes, but I left the rest of it up to you – 2017-01-25