Hard intergral

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.

 

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.

 

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)

 

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)?

 

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.