# Thread: is this derivative correct?

1. ## is this derivative correct?

y = e^-0.4x(C cos0.2x + Dsin0.2x)
y' = -0.4e^-0.4x(-0.2C sin0.2x + 0.2 D cos0.2x)

This from trying to work out a particular solution to a inhomogeneous equation

Paul B

2. Originally Posted by poundedintodust
y = e^-0.4x(C cos0.2x + Dsin0.2x)
y' = -0.4e^-0.4x(-0.2C sin0.2x + 0.2 D cos0.2x)

This from trying to work out a particular solution to a inhomogeneous equation

Paul B
You have to follow the product rule.

$\displaystyle y = e^{-0.4x}[C cos(0.2x) + Dsin(0.2x)]$

$\displaystyle y'=-0.4e^{-0.4x}[C cos(0.2x) + Dsin(0.2x)]+e^{-0.4x}[0.2Dcos(0.2x)-0.2Csin(0.2x)]$

3. if y = 3 and x=0

should i be left with

0.4C + 0.2D = 3?

4. Originally Posted by poundedintodust
if y = 3 and x=0

should i be left with

0.4C + 0.2D = 3?
If $\displaystyle y(0)=3$, then $\displaystyle C=3$

5. I am a bit confused now so i think Ill post whole problem and maybe you could tell me where I am going wrong.

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