1. ## Differential Equations

The question states to find the solution of the following -

$\displaystyle \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 2e^x$

Ive got quite far, but when trying to find the value of 'C' i keep getting this:

C-2C+C=2

Which gives no value for C, any help will be appreciated, as i really want to finish this question..

2. Hello, brd_7!

I don't know what "C" you're referring to . . .

Find the solution of: .$\displaystyle \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y \:= \;2e^x$

The characteristic equation is: .$\displaystyle m^2-2m + 1\:=\:0\quad\Rightarrow\quad (m-1)^2\:=\:0$

. . which has roots: .$\displaystyle m \;=\;1,\,1$

The homogeneous solution is: .$\displaystyle y \;=\;C_1e^x + C_2xe^x$

Then the particular solution is of the form: .$\displaystyle y_p \;=\;Ax^2e^x$

. . and (by your favorite method) we find that: .$\displaystyle y_p \:=\:x^2e^x$

Therefore, the solution is: .$\displaystyle \boxed{y \;=\;C_1e^x + C_2xe^x + x^2e^x}$

Without further information, we cannot evaluate the C's.

3. Why do you use:
$\displaystyle y_p \;=\;Ax^2e^x$
And not:
$\displaystyle y_p \;=\;Ae^x$

4. Originally Posted by brd_7
Why do you use:
$\displaystyle y_p \;=\;Ax^2e^x$
And not:
$\displaystyle y_p \;=\;Ae^x$
Because $\displaystyle Ae^x$ is part of the homogenous solution (try using it and solving for A. Uh oh ....). So you might try the next simplest, which is $\displaystyle Axe^x$ but uh oh ...... So try the next simplest, which is .....

5. Ah right, ok, because i used Ae^x and it didnt work, which is where i got suck originally, i never knew you could do that, now i do, im very thankful!