# Thread: Last question of the day on differential equations

1. ## Last question of the day on differential equations

So I have this problem which states use separation of variables to find the general solution of the differential equation.

$\frac{dy}{dx}=\sqrt\frac{x}{y}$ so to start, I'll move the dx to the other side to get:

$dy=\sqrt\frac{x}{y}dx$

Now, to get rid of the square root, can I simply square both sides to get:

$d^{2}y=\frac{x}{y}dx^2$ then divide by y to get:

$\frac{1}{y}d^{2}y=xdx^{2}$

And then from there, integrate both sides twice? The only issue I see is after integrating once, I get:

$lnydy=\frac{1}{2}x^2dx+c$

And I don't quite remember how to integrate a natural log...

2. $\frac{dy}{dx}=\sqrt{\frac{x}{y}}$

$\frac{dy}{dx}=\frac{\sqrt{x}}{\sqrt{y}}$

$\sqrt{y}~dy = \sqrt{x}~dx$

Now integrate..

3. Originally Posted by emttim84
So I have this problem which states use separation of variables to find the general solution of the differential equation.

$\frac{dy}{dx}=\sqrt\frac{x}{y}$ so to start, I'll move the dx to the other side to get:

$dy=\sqrt\frac{x}{y}dx$

Now, to get rid of the square root, can I simply square both sides to get:

$d^{2}y=\frac{x}{y}dx^2$ then divide by y to get:

$\frac{1}{y}d^{2}y=xdx^{2}$

And then from there, integrate both sides twice? The only issue I see is after integrating once, I get:

$lnydy=\frac{1}{2}x^2dx+c$

And I don't quite remember how to integrate a natural log...
And just for you knowledge using integration by parts $\int\ln(ax)dx=x\cdot{ln(ax)}-\int{\frac{x\cdot{a}}{a\cdot{x}}}=x\cdot{ln(ax)}-x+C$

4. ....well, don't I feel like a retard. And now you know why I'm not a math major. :P thanks guys.

5. Originally Posted by emttim84
....well, don't I feel like a retard. And now you know why I'm not a math major. :P thanks guys.
Haha that is why we are here!