Homework Help: Hard intergral

1. Mar 2, 2010

Sirsh

Hard intergral!!

If y = ke-x(sin2x), prove that $$\frac{d^2y}{dx^2}$$ + 2$$\frac{dy}{dx}$$ + 5y = 0

Need help asap, No clue how to figure it out. teacher is away until next term, test next week.

2. Mar 2, 2010

rock.freak667

Re: Hard intergral!!

you have y, you can find d2y/dx2 and dy/dx. Just find them and sub them into the equation and see if the left side is the same as the right side.

3. Mar 2, 2010

Sirsh

Re: Hard intergral!!

y'= sin2x(-ke-x)-(sin2x*cos2x)

y''= ke-xcos2x+sin2x(ke-x)+1

I've got my first and second intergral, can i get a confirmation? thanks alot.

4. Mar 2, 2010

Dick

Re: Hard intergral!!

No, not right. All of your terms in y' and y'' should have a k and exp(-x) in them. Do you see why? Just use the product rule and the chain rule.

5. Mar 2, 2010

Sirsh

Re: Hard intergral!!

I dont see why there would be a exp(-x)?? i thought the derivative of e^x was e^x

6. Mar 2, 2010

cronxeh

Re: Hard intergral!!

Derivative of exp(-x) is -exp(-x)

But more importantly you have 2 functions that you are taking a derivative of. Also, stop saying 'integral' - that is not what you are doing.

And finally,

(f*g)' = f' * g + f * g'

y=k*exp(-x)*sin(2*x)

f=k*exp(-x)
g=k*sin(2*x)

y=k*(f*g)

Find y' = (f*g)' = f' * g + f * g'

when f' = -exp(-x)
and g'=2*cos(2*x)

Last edited: Mar 2, 2010
7. Mar 2, 2010

Dick

Re: Hard intergral!!

Because the problem you started out with has exp(-x) in it, not exp(x). What's the derivative of exp(-x)?

8. Mar 2, 2010

Sirsh

Re: Hard intergral!!

lol sorry i put integral.. i am also doing integrals at the moment, kind of side tracked.
alright so the frist derivative is -e^(-x)sin2x+ke^(-x)2cos2x.
And second derivative is (e^(-x)2cos2x-e^(-x)sin2x)+(-ke^(-x)2cos2x-ke^(-x)4sin2x)

9. Mar 2, 2010

Dick

Re: Hard intergral!!

That's getting better. But why no k in the first term of y'=-e^(-x)sin2x+ke^(-x)2cos2x? I told you there should be one.