Second order differential equation

[Tan Bee Yong]

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

 

[tiny-tim]

Welcome to PF!

Hi Tan Bee Yong! Welcome to PF!

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

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

 

[Tan Bee Yong]

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 don't understant. why is there an x?, i thought p should be a constant)

sori i don't know how to use the button to show ^

 

[Tan Bee Yong]

Y_p = Ax + B + pxe^2x
Y_p^' : A + pe^2x + 2pxe^2x
Y_p^'' 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 don't understant. why is there an x?, i thought p should be a constant)

 

[tiny-tim]

Hi Tan Bee Yong!

(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^2x …

your particular solution should therefore be Ax + B + Ce^2x.

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

You only need to use xe^2x if e^2x is a solution to the LHS (which in this case, it isn't!).

 

[Tan Bee Yong]

LHS has a pair of complex conjugate roots, hence my Y_c = e^2x(Dcos(x) + Ecos(x))

i thought this is a special case, that's why I use Cxe^2x

Did I make a mistake at the complementary function?

 

[tiny-tim]

No, the roots are 2 ± i, that's not the same as 2 !

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

 

[Tan Bee Yong]

No, the roots are 2 ± i, that's not the same as 2 !

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