# Proving the general solution of a differential equation

• Dec 18th 2011, 11:49 PM
Punch
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?
• Dec 19th 2011, 12:07 AM
FernandoRevilla
Re: Proving the general solution of a differential equation
Quote:

Originally Posted by Punch
$\displaystyle \frac{1}{2}y^2=Alnx+C$ how can I twit the equation further?

$\displaystyle y^2=2A\ln x+2C=\ln x^{2A}+\ln K=\ln (Kx^{2A})=\ln (Kx^n)$
• Dec 19th 2011, 12:09 AM
Punch
Re: Proving the general solution of a differential equation
Quote:

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
• Dec 19th 2011, 12:10 AM
Punch
Re: Proving the general solution of a differential equation
Quote:

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!