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What is the general solution of differential equation $y\frac{d^{2}y}{dx^2} - (\frac{dy}{dx})^2 = y^2 log(y)$.

The answer to this DE is $log(y) = c_1 e^x + c_2 e^{-x}$

I don't know the method to solve differential equation with degree more than 1. Please tell me how to solve these types of equations.

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    $\begingroup$ Is it $\log(x)$ or $\log(y)$ on the right side? Your equation transforms to $(\log(y))''=\log(x)$ which integrates differently. $\endgroup$ – Dr. Lutz Lehmann Apr 30 at 4:24
  • $\begingroup$ You are right, the answer given must be wrong, I will change it. $\endgroup$ – sawan kumawat Apr 30 at 4:52
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    $\begingroup$ Now with $u=\log(y)$ you get the simple linear DE $u''=u$ or $u''-u=0$ which has indeed the proposed solution. $\endgroup$ – Dr. Lutz Lehmann Apr 30 at 5:18
  • $\begingroup$ Yo can post it as an answer. $\endgroup$ – sawan kumawat Apr 30 at 5:25
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Let's suppose $y\neq 0$, then the given DE is equivalent to $$\dfrac{y\dfrac{d^2y}{dx^2}-\left(\dfrac{dy}{dx}\right)^2}{y^2}=\log y$$ i.e. $$\dfrac{d}{dx}\left(\dfrac1y\dfrac{dy}{dx}\right)=\log y$$ By making the sustitution $u=\log y$ the last DE becomes $$\dfrac{d}{dx}\left(\dfrac{du}{dx}\right)=u\qquad \text{i.e.} \qquad u''-u=0$$ last equation is linear and homogeneous, its solution is given by $u=c_1e^x+c_2e^{-x}$.

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  • $\begingroup$ I am getting $log(y) = \frac{x^{2}log(x)}{2} - \frac{3x^2}{4}$ but this is not in the form of $c_1 e^x + c_2 e^{-x}$ $\endgroup$ – sawan kumawat Apr 30 at 5:06
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Hint. The equation can be written as $$\frac{d}{dx}\Bigl(\frac{\frac{dy}{dx}}{y}\Bigr)=\log x\ .$$ There is in general no specific procedure for solving this kind of thing and you have to rely on "spotting" something like the above.

BTW I don't think the answer you have given is correct. Probably the $\log x$ should be $\log y$ as suggested by LutzL.

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