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
$\displaystyle \frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 16y = 2e^{4x}$

My answer:

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

Does this look correct? Thanks.

Quote:

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Originally Posted by Glitch

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

My answer:

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

Does this look correct? Thanks.

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$\displaystyle y''-8y'+16y=0$

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

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

Quote:

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Originally Posted by dwsmith

$\displaystyle y''-8y'+16y=0$

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

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

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

Quote:

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

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I have selective reading. I only read homogeneous and not non-homogeneous.

Sorry, my title was probably misleading. Thanks guys.