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$$y'=y^2 \ln x+\frac{y}{x}$$

This is Bernoulli equation of the form $y'+p_{1}(x)y=p_{2}(x)y^n$

So we will multiply the equation by $y^{-2}$ to get:

$$y^{-2}y'=\ln x+\frac{1}{x}y^{-1}$$

$$y^{-2}y'-\frac{1}{x}y^{-1}=\ln x$$

Let $t=y^{-1}\text{ , } t'=-y^{-2}y'$

So we have

$$t'+\frac{1}{x}t=-\ln x$$

Now I can not get rid of $\frac1x$ is it because the ode is non linear? Why is that? It seems that all the operations on the $y$'s are linear functions.

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    The ODE $$t' + \frac{1}{x} t = -\ln x$$ is linear. – 2017-01-30

3 Answers 3

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Hint: The ODE is linear. It only has nonconstant coefficients. the general solution is given as the superposition of the homogeneous solution and the particular solution.

The homogeneous equaton is: $t'_h+\frac{1}{x}t_h =0.$ This homogeneous equation (Euler-Cauchy-ODE) can be solved by a power ansatz: $t_h=ax^n$. The particular solution $t_p$ can be obtained by variation of parameters.

The general solution is: $t=t_h+t_p$.

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    So there is no need to use integrating factor? when will integrating factor must be used? – 2017-01-30
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    Integrating factor should also work. It is even simpler. – 2017-01-30
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If you multiply the equation by $x$, the left side is $xt'+t$ which is the derivative of $xt$.

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@gbox: by forwarding your answer $$t'+\frac{1}{x}t=-\ln x$$,which is a linear equation.Now Integrating factor(IF):$e^{\int\frac1x dx}=e^{\log x}=x$.Multiplying it in above equation we have $$xt'+t=-x\ln x$$ $$\frac{d}{dx}xt=-x\ln x$$ $$xt=\int-x\ln x dx=\frac14x^2(1-2\log x)+c$$.Now put $t=y^{-1}$,then our solution is given by $$\frac xy=\frac14x^2(1-2\log x)+c$$