Which is the correct expression for the Differential Equation

[Northbysouth]

Homework Statement

Which of the following is a correct expression for

∫₁^t(s-1)y'(s) ds

I know the answer is:

a) 1/2(t-1)²y(t)

b) 1/2(t-1)²y'(t)+ (t-1)(y(t)-y(1))

c) (t-1)(y(t)-y(1))

d) (t-1)y(t)-∫y(s) ds

Homework Equations

The Attempt at a Solution

I know the answer is d, but I don't understand why. Are they integrating the given equation, and if so, how do you integrate a function without knowing what the function is?

I have attached an image of the original question

 

[mfb]

You can do integration by parts - the remaining integral is just an integral over y(s), and you can find that in answer (d). The other part can be evaluated without an integral.

 

[Dick]

Homework Statement

Which of the following is a correct expression for

∫₁^t(s-1)y'(s) ds

I know the answer is:

a) 1/2(t-1)²y(t)

b) 1/2(t-1)²y'(t)+ (t-1)(y(t)-y(1))

c) (t-1)(y(t)-y(1))

d) (t-1)y(t)-∫y(s) ds

Homework Equations

The Attempt at a Solution

I know the answer is d, but I don't understand why. Are they integrating the given equation, and if so, how do you integrate a function without knowing what the function is?

I have attached an image of the original question

You'll notice they didn't actually integrate it in choice d). The just expressed the original integral in terms of another integral. The technique is called 'integration by parts'. It should be described in your course.

 

[Northbysouth]

I think I've got it now. It's been a while since I've done integration by parts. Thanks for the input.