# SOS! Proving a Hard Integral Before Next Week's Test

• Sirsh
In summary: Also, don't forget that you'll need to use the product rule again to find y''.In summary, the problem is asking to prove that y'' + 2y' + 5y = 0 for the function y = ke-x(sin2x). To do this, you must first find the first and second derivatives of y using the product rule and the chain rule. After finding these derivatives, you can substitute them into the equation and simplify to see if the left side equals the right side. Remember to include the k and exp(-x) terms in all of your derivatives.
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.

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.

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.

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.

I don't see why there would be a exp(-x)?? i thought the derivative of e^x was e^x

Sirsh said:
I don't 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)

Last edited:

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

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)

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.

## 1. How do I solve a hard integral quickly for an upcoming test?

Solving a hard integral quickly requires knowledge of various techniques such as substitution, integration by parts, and trigonometric identities. It is important to practice these techniques beforehand and approach the integral systematically to save time.

## 2. What are some common mistakes to avoid when solving a hard integral?

Some common mistakes to avoid when solving a hard integral include forgetting to use the chain rule, not simplifying the expression before integrating, and making sign errors. It is important to double check your work and be mindful of these mistakes.

## 3. How can I break down a hard integral into smaller, more manageable parts?

Breaking down a hard integral into smaller parts can be done by using techniques such as partial fraction decomposition or splitting the integral into multiple integrals. This can make the problem more approachable and easier to solve.

## 4. What are some resources I can use to help me solve a hard integral?

There are various resources available to help you solve a hard integral, such as online integral calculators, textbooks, and tutoring services. It is also helpful to work with classmates or seek assistance from your professor if you are struggling.

## 5. How can I practice solving hard integrals before my upcoming test?

The best way to practice solving hard integrals is to work through a variety of problems and to challenge yourself with more difficult ones. You can find practice problems in your textbook, online resources, or ask your professor for additional practice materials. It is important to also review and understand any mistakes you make in the process.

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