Second order differential equation

  • #1

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.


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
tiny-tim
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Homework Helper
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Welcome to PF!

Hi Tan Bee Yong! Welcome to PF!

(try using the X2 icon just above the Reply box :wink:)

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
  • #3
particular solution = Ax + B + pxe^2x
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
Yp = Ax + B + pxe2x
Yp' : A + pe^2x + 2pxe2x
Yp'' 2pxe2x + 2pxe2x + 4pxe2x=4pe2x+ 4pxe2x

substitute the above into the DE. i get:

4pe2x + 4pxe2x -4A-4pe2x - 8pxe2x + 5Ax + 5B + 5pxe2x

=-4A + 5Ax + 5B + pxe2x

by comparing with the right-hand side of DE;

5Ax = 25x
A= 5,

-4A + 5B = 0
B = 4

pxe2x = -3e2x
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
Science Advisor
Homework Helper
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251
Hi Tan Bee Yong! :smile:

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

Your original RHS was 25 - 3e2x

your particular solution should therefore be Ax + B + Ce2x.

(btw, why use capital A and B and small p? it's not illegal, but I think C is neater! :wink:)

You only need to use xe2x if e2x is a solution to the LHS (which in this case, it isn't!). :smile:
 
  • #6
LHS has a pair of complex conjugate roots, hence my Yc = e2x(Dcos(x) + Ecos(x))

i thought this is a special case, that's why I use Cxe2x

Did I make a mistake at the complementary function?
 
  • #7
tiny-tim
Science Advisor
Homework Helper
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251
No, the roots are 2 ± i, that's not the same as 2 ! :wink:

You only need x if the RHS exponent is exactly the same as a LHS root.
 
  • #8
No, the roots are 2 ± i, that's not the same as 2 ! :wink:

You only need x if the RHS exponent is exactly the same as a LHS root.
Ok .. will give it another try.. thank you so much for helping
 

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