Need a general solution checked

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February 5th 2011, 07:51 PM
Glitch

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

My answer:

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

Does this look correct? Thanks.
February 5th 2011, 07:55 PM
dwsmith

Quote:

Originally Posted by Glitch

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

My answer:

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

Does this look correct? Thanks.

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

$(m-4)^2=0$

$y_p=C_1e^{4x}+C_2xe^{4x}$
February 5th 2011, 08:09 PM
Prove It

Quote:

Originally Posted by dwsmith

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

$(m-4)^2=0$

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

What you've written is the Characteristic Solution, not the Particular Solution.

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

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

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

So I agree with the OP's choice of $\displaystyle y_p = Cx^2e^{4x}$ .
February 5th 2011, 08:11 PM
dwsmith

Quote:

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 $\displaystyle e^{4x}$ , you would normally choose $\displaystyle Ce^{4x}$ as a particular solution.

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

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

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

I have selective reading. I only read homogeneous and not non-homogeneous.
February 5th 2011, 08:40 PM
Glitch

Sorry, my title was probably misleading. Thanks guys.