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The question is to solve $x\frac{\mathrm{d} y}{\mathrm{d} x}+3y=x^3y^2$.

I have found the homogenous solution $y_h = c_1 x^{-3}$

I am stuck at finding the particular solution. I am familiar with variation of parameters(which involve Wronskian and just $r(x)$ in RHS ) and solution by undetermined coefficients

Please do as well suggest any necessary reading required for the same

Soham

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    The original equation is the form of [Bernoulli's equation](http://m.cliffsnotes.com/study_guide/Bernoullis-Equation.topicArticleId-19736,articleId-19715.html). This suggests the substitution $w=y^{1-3}$ to reduce the equation to linear in $w$.2012-09-12
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    Much thanks. I had been at my wits end, and now I realize the answer was just there under my nose. Much thanks.2012-09-12
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    Voted to close it2012-09-12
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    Why? There's nothing wrong with question--it's well-written, polite and shows effort. If only every question here was like this one.2012-10-19

2 Answers 2

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You can probably use an integrating factor. Multiply by $x^2/y^2$, and after some rearrangements you get

$$\eqalign{ & xy' + 3y = {x^3}{y^2} \cr & 3{x^2}\frac{1}{y} - {x^3}\left( { - \frac{{y'}}{{{y^2}}}} \right) = {x^5} \cr & \frac{d}{{dx}}\left( {{x^3}} \right)\frac{1}{y} - {x^3}\frac{d}{{dx}}\left( {\frac{1}{y}} \right) = {x^5} \cr} $$

Now, divide by $x^6$ and negate to get

$$ - \frac{{\frac{d}{{dx}}\left( {\frac{1}{y}} \right){x^3} - \frac{d}{{dx}}\left( {{x^3}} \right)\frac{1}{y}}}{{{x^6}}} = \frac{1}{x}$$

$$ - \frac{d}{{dx}}\left( {\frac{1}{{{yx^3}}}} \right) = \frac{1}{x}$$

So that $$\eqalign{ & C - \frac{1}{{{yx^3}}} = \log x \cr & y = \frac 1 {{x^3}\left( {C - \log x} \right)} \cr} $$

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I assume there's a general method in solving equations like this, but I'm a little rusty. I do think I see a way to solve this though. First, we'll multiply by an integrating factor of $x^2$.

$$x^3\frac{dy}{dx}+3x^2y=x^5y^2$$

Now make the substitution

$$z=x^3y,\frac{dz}{dx}=x^3\frac{dy}{dx}+3x^2y$$

$$\frac{dz}{dx}=\frac{z^2}{x}$$

We now have a separable equation.

$$\frac{dz}{z^2}=\frac{dx}x$$

$$-\frac1z=\ln x+c$$

$$z=\frac1{c_1-\ln x}$$

$$y=\frac1{x^3(c_1-\ln x)}$$