Last question of the day on differential equations

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April 19th 2008, 12:52 PM
emttim84

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...
April 19th 2008, 01:07 PM
wingless

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

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

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

Now integrate..
April 19th 2008, 03:05 PM
Mathstud28

Quote:

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 $\int\ln(ax)dx=x\cdot{ln(ax)}-\int{\frac{x\cdot{a}}{a\cdot{x}}}=x\cdot{ln(ax)}-x+C$
April 19th 2008, 06:21 PM
emttim84

....well, don't I feel like a retard. And now you know why I'm not a math major. :P thanks guys.
April 19th 2008, 06:35 PM
Mathstud28

Quote:

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!