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

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.

• Is it $\log(x)$ or $\log(y)$ on the right side? Your equation transforms to $(\log(y))''=\log(x)$ which integrates differently. – Dr. Lutz Lehmann Apr 30 at 4:24
• You are right, the answer given must be wrong, I will change it. – sawan kumawat Apr 30 at 4:52
• Now with $u=\log(y)$ you get the simple linear DE $u''=u$ or $u''-u=0$ which has indeed the proposed solution. – Dr. Lutz Lehmann Apr 30 at 5:18
• Yo can post it as an answer. – sawan kumawat Apr 30 at 5:25

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 $$u=c_1e^x+c_2e^{-x}$$.
• I am getting $log(y) = \frac{x^{2}log(x)}{2} - \frac{3x^2}{4}$ but this is not in the form of $c_1 e^x + c_2 e^{-x}$ – sawan kumawat Apr 30 at 5:06
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.