# Thread: Need a general solution checked

1. ## Need a general solution checked

The question:
Write down the form of the particular solution you should try when solving the non-homogeneous differential equation
$\displaystyle \frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 16y = 2e^{4x}$

$\displaystyle y_p = Cx^2e^{4x}$

Does this look correct? Thanks.

2. Originally Posted by Glitch
The question:
Write down the form of the particular solution you should try when solving the non-homogeneous differential equation
$\displaystyle \frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 16y = 2e^{4x}$

$\displaystyle y_p = Cx^2e^{4x}$

Does this look correct? Thanks.
$\displaystyle y''-8y'+16y=0$

$\displaystyle (m-4)^2=0$

$\displaystyle y_p=C_1e^{4x}+C_2xe^{4x}$

3. Originally Posted by dwsmith
$\displaystyle y''-8y'+16y=0$

$\displaystyle (m-4)^2=0$

$\displaystyle y_p=C_1e^{4x}+C_2xe^{4x}$
What you've written is the Characteristic Solution, not the Particular Solution.

Since your RHS of the DE is of the family $\displaystyle \displaystyle e^{4x}$, you would normally choose $\displaystyle \displaystyle Ce^{4x}$ as a particular solution.

But since $\displaystyle \displaystyle e^{4x}$ appears in your Characteristic Solution, you would then normally choose $\displaystyle \displaystyle Cx\,e^{4x}$ as a particular solution.

But since $\displaystyle \displaystyle x\,e^{4x}$ appears in your Characteristic Solution, you would then normally choose $\displaystyle \displaystyle Cx^2e^{4x}$ as a particular solution.

So I agree with the OP's choice of $\displaystyle \displaystyle y_p = Cx^2e^{4x}$.

4. Originally Posted by Prove It
What you've written is the Characteristic Solution, not the Particular Solution.

Since your RHS of the DE is of the family $\displaystyle \displaystyle e^{4x}$, you would normally choose $\displaystyle \displaystyle Ce^{4x}$ as a particular solution.

But since $\displaystyle \displaystyle e^{4x}$ appears in your Characteristic Solution, you would then normally choose $\displaystyle \displaystyle Cx\,e^{4x}$ as a particular solution.

But since $\displaystyle \displaystyle x\,e^{4x}$ appears in your Characteristic Solution, you would then normally choose $\displaystyle \displaystyle Cx^2e^{4x}$ as a particular solution.

So I agree with the OP's choice of $\displaystyle \displaystyle y_p = Cx^2e^{4x}$.