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## Homework Statement

This question concerns the second-order differential equation

dy^2/dx^2 -4(dy/dx) + 5y = 25x - 3e^2x. Find the solution of the differential equation that satisfies the initial conditions y=0 and dy/dx= 0 when x=0.

- Thread starter Tan Bee Yong
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- #1

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This question concerns the second-order differential equation

dy^2/dx^2 -4(dy/dx) + 5y = 25x - 3e^2x. Find the solution of the differential equation that satisfies the initial conditions y=0 and dy/dx= 0 when x=0.

- #2

tiny-tim

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Hi Tan Bee Yong! Welcome to PF!

(try using the X

Show us what you've tried, and where you're stuck, and then we'll know how to help!

- #3

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differential once : A + pe^2x + 2pxe^2x

differential again 2pxe^2x + 2pxe^2x + 4pxe^2x=4pe^2x + 4pxe^2x

substitute the above into the DE. i get:

4pe^2x + 4pxe^2x -4A-4pe^2x - 8pxe^2x + 5Ax + 5B + 5pxe^2x

=-4A + 5Ax + 5B + pxe^2x

by comparing with the right-hand side of DE;

5Ax = 25x

A= 5,

-4A + 5B = 0

B = 4

pxe^2x = -3e^2x

px=-3

p= -3/x (this is the part i dont understant. why is there an x?, i thought p should be a constant)

sori i dont know how to use the button to show ^

- #4

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Y

Y

substitute the above into the DE. i get:

4pe

=-4A + 5Ax + 5B + pxe

by comparing with the right-hand side of DE;

5Ax = 25x

A= 5,

-4A + 5B = 0

B = 4

pxe

px=-3

p= -3/x (this is the part i dont understant. why is there an x?, i thought p should be a constant)

- #5

tiny-tim

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(if you click the "QUOTE" button, that takes you to the Reply page, and there's a row of icons above the reply box )

Your original RHS was 25 - 3e

your particular solution should therefore be Ax + B + Ce

(btw, why use capital A and B and small p? it's not

You only need to use xe

- #6

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i thought this is a special case, that's why I use Cxe

Did I make a mistake at the complementary function?

- #7

tiny-tim

Science Advisor

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You only need x if the RHS exponent is

- #8

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Ok .. will give it another try.. thank you so much for helpingnotthe same as 2 !

You only need x if the RHS exponent isexactlythe same as a LHS root.

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