# Thread: Proving the general solution of a differential equation

1. ## Proving the general solution of a differential equation

Prove that the general solution to the differential equation, $\displaystyle \frac{dy}{dx}=\frac{k}{xhy}$ is $\displaystyle y^2=lnCx^n$ where $\displaystyle k, h, C and n$ are constants.

$\displaystyle \frac{dy}{dx}=\frac{k}{xhy}$

$\displaystyle \int{y}dy=\int\frac{A}{x}dx$ where $\displaystyle A=\frac{k}{h}$

$\displaystyle \frac{1}{2}y^2=Alnx+C$

how can I twit the equation further?

2. ## Re: Proving the general solution of a differential equation

Originally Posted by Punch
$\displaystyle \frac{1}{2}y^2=Alnx+C$ how can I twit the equation further?
$\displaystyle y^2=2A\ln x+2C=\ln x^{2A}+\ln K=\ln (Kx^{2A})=\ln (Kx^n)$

3. ## Re: Proving the general solution of a differential equation

Originally Posted by FernandoRevilla
Express the equation in the form $\displaystyle hy\;dy-\frac{kdx}{x}=0$ (separated variables) .
Alright, then i get $\displaystyle h\frac{y^2}{2}=klnx + C$ which I still dont see how I can further transform this

4. ## Re: Proving the general solution of a differential equation

Originally Posted by FernandoRevilla
$\displaystyle y^2=2A\ln x+2C=\ln x^{2A}+\ln K=\ln (Kx^{2A})=\ln (Kx^n)$
Oh! I see, so I just have to insert a lnC' to replace C and that does the trick!