# particular soultion of a initial value problem

• Mar 4th 2008, 09:42 AM
poundedintodust
particular soultion of a initial value problem
i know this is more than likely a calculus problem but not sure if my algebra is right so hence the re posting over here. If its the wrong way to post it sorry!

Starting with:

5(d^2y/d^2x) + 4(dy/dx) + y = 0

I have find a particular solution to the initial value problem where

y(0) = -2, y'(0)=3

I know a general solution is

y = e^-0.4x(C cos0.2x + Dsin0.2x)

after finding the roots.

So substituting in the values into the derivative

y' = -0.4e^-0.4x(-0.2C sin0.2x + 0.2 D cos0.2x)

ends with 0.4C+0.2D =3

So C = 3 and D = 9

Doing the same with the general solution with x=-2

y = e^-0.4x(C cos0.2x + Dsin0.2x)

I end up with 1( C x 1 + D x 0) in other words C = -2

My confusion comes from the fact that C should be the same in both instances??

Cheers for the help
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• Mar 4th 2008, 12:31 PM
Opalg
Quote:

Originally Posted by poundedintodust
Starting with:

5(d^2y/d^2x) + 4(dy/dx) + y = 0

I have find a particular solution to the initial value problem where

y(0) = -2, y'(0)=3

I know a general solution is

y = e^-0.4x(C cos0.2x + Dsin0.2x)

after finding the roots. (I haven't checked that, so I'll assume it's correct.)

So substituting in the values into the derivative

y' = -0.4e^-0.4x(-0.2C sin0.2x + 0.2 D cos0.2x)

ends with 0.4C+0.2D =3 No! What you are told is that y'(0)=3. This means that you put x=0 in the expression for y', which tells you that -0.4(0.2 D)=3 (since e^0 and cos0 are both 1, and sin0 is 0).

So C = 3 and D = 9

Doing the same with the general solution with x=-2 That should be y=-2 when x=0

y = e^-0.4x(C cos0.2x + Dsin0.2x)

I end up with 1( C x 1 + D x 0) in other words C = -2 This time you've got it right!

My confusion comes from the fact that C should be the same in both instances??

..