Homework Help: Hard intergral

Sirsh · Mar 2, 2010

Hard intergral!!

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

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

rock.freak667 · Mar 2, 2010

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.

Sirsh · Mar 2, 2010

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.

Dick · Mar 2, 2010

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.

Sirsh · Mar 2, 2010

Re: Hard intergral!!

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

cronxeh · Mar 2, 2010

Re: Hard intergral!!

Sirsh said: ↑

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

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)

Dick · Mar 2, 2010

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)?

Sirsh · Mar 2, 2010

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)

Dick · Mar 2, 2010

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.