# Linear first-order equation Problem

• Jun 13th 2009, 07:01 PM
noppawit
Linear first-order equation Problem
I am quite not sure about my solution to solve this by using Linear equation method.

$\displaystyle y\frac{dx}{dy}-x = 2y^2 ; y(1) = 5$

Then, to follow the form $\displaystyle \frac{dy}{dx}+P(x)y=f(x)$ I get $\displaystyle \frac{dx}{dy}-(\frac{1}{y})x=2y$ (Because I assume, x=y and y=x)
Integrating factor is $\displaystyle e^{\int\frac{1}{y}dy} = e^{ln|y|} = y$
After that, I solved:
$\displaystyle \frac{d}{dy}(xy) = 2y^2$
$\displaystyle xy = \frac{2y^3}{3}+C1$
$\displaystyle x = \frac{2y^2}{3}+C2$

After I have done like this, I don't know what should I do with y(1) = 5.

• Jun 13th 2009, 11:44 PM
Jhevon
Quote:

Originally Posted by noppawit
I am quite not sure about my solution to solve this by using Linear equation method.

$\displaystyle y\frac{dx}{dy}-x = 2y^2 ; y(1) = 5$

Then, to follow the form $\displaystyle \frac{dy}{dx}+P(x)y=f(x)$ I get $\displaystyle \frac{dx}{dy}-(\frac{1}{y})x=2y$ (Because I assume, x=y and y=x)
Integrating factor is $\displaystyle e^{\int\frac{1}{y}dy} = e^{ln|y|} = y$
After that, I solved:
$\displaystyle \frac{d}{dy}(xy) = 2y^2$
$\displaystyle xy = \frac{2y^3}{3}+C1$
$\displaystyle x = \frac{2y^2}{3}+C2$

you have the right idea, however, you should have realized that your integrating factor is off. $\displaystyle \frac d{dy}(xy)$ does not yield $\displaystyle y \frac {dx}{dy} - x$ as it should. rather your integrating factor is given by $\displaystyle e^{\int - \frac 1ydy}$

Quote:

After I have done like this, I don't know what should I do with y(1) = 5.
y(1) = 5 simply means when x = 1, y = 5. you are supposed to plug in these values into your solution (after you correct it :p) and this will allow you to solve for the arbitrary constant $\displaystyle C_2$