1. Equations

Hello, was just wondering if someone could give me some help.
i have to find the solution of the equation
y'-2xy=x which satisfies y(0)=0

so far i have done the basic and written y'= 2xy + x
next i was going to divide all through by y but got a bit stuck after that point. it probably is a relatively easy question but i just am not sure where to go next... if anyone could lend a hand it would be appreciated!

2. Originally Posted by biffer
Hello, was just wondering if someone could give me some help.
i have to find the solution of the equation
y'-2xy=x which satisfies y(0)=0

so far i have done the basic and written y'= 2xy + x
next i was going to divide all through by y but got a bit stuck after that point. it probably is a relatively easy question but i just am not sure where to go next... if anyone could lend a hand it would be appreciated!
Have you tried an integrating factor?

-Dan

3. thanks!

4. You can also try solving the equation without integrating factor..

$\displaystyle y' - 2 x y = x$

First solve the LHS..

$\displaystyle y' - 2 x y = 0$

$\displaystyle y' = 2 x y$

$\displaystyle \frac{dy}{y} = 2x~dx$

$\displaystyle \int \frac{dy}{y} = \int 2x~dx$

$\displaystyle ln|y|=x^2+C$

$\displaystyle y=C\cdot e^{x^2}$....$\displaystyle \boxed{I}$

Now take C as a function, and find y'

$\displaystyle y' = (C')(e^{x^2}) + (C)(2xe^x)$

$\displaystyle y' = C'e^{x^2} + 2Cxe^x$....$\displaystyle \boxed{II}$

Put y and y' (marked as I and II) in the original equation:

$\displaystyle y' - 2 x y = x$

$\displaystyle C'e^{x^2} + 2Cxe^x - 2 x C e^{x^2} = x$

$\displaystyle C'e^{x^2} = x$

$\displaystyle dC = \frac{x}{e^{x^2}}~dx$

$\displaystyle C = -\frac{1}{2e^{-x^2}} + K$

Now put this C in I,

$\displaystyle y=Ce^{x^2}$

$\displaystyle y=(-\frac{1}{2e^{-x^2}} + K)e^{x^2}$

$\displaystyle y= -\frac{1}{2} + K\cdot e^{x^2}$