Efficient Integration Techniques: Solving for y' and Finding the Integral

[MustangGt94]

Help with Integration :(

Ok guys I've been trying to solve this problem for a couple of days but no luck :( Please help me out! I think I have part of it done.

I've added the problem as an attachment wasn't sure how to type it up properly. But when i solve for y' i get the answer 2t+e^(t^2) I am however having trouble solving the integral in order to get a value for y to plug into the equation.

Thanks for the help!

 

[sutupidmath]

i am confused here, what are you really trying to do?

I mean, solve for y, or t, at the end?

It would be a good idea to post the entire problem, with the text, so it will clarify things.

 

[MustangGt94]

Oh my bad here is the problem:

Verify that each given function is a solution of the differential equation. So I was assuming to solve for y' and plug it into the initial equation and check to see if the condition equals 1.

 

[Defennder]

You are asked only to verify that the expression for y(t) satisfies the DE y'-2ty=1. If you work out what is y' (using the fundamental theorem of calculus) and substitute it into the DE, you'll find that it all cancels out nicely to give you 1 as expected.

 

[MustangGt94]

Thats where i am running into some problems defender. I solved for y' and i got 2t+e^(t^2) was wondering if someone can check that. But also do I have to solve the integral in y before I can plug it back into the original DE? Those are the step that I am taking so far but I am having trouble solving the integral in y in order to plug it back into the DE. :(

 

[Defennder]

There isn't any way to evalute the integral in elementary functions and you do not have to. Your expression for y' should have the integral inside. I don't know how you got that but you have to use the product rule for differentiation.

 

[HallsofIvy]

You are given, I suppose,
y= e^{t^2}\int_0^t e^{-s^2}ds+ e^{t^2}
I think I would be inclined first to write it as
y= e^{t^2}\left(\int_0^t e^{-s^2}ds+ 1\right)

Now, to see if that function satisfies the differential equation differentiate that to find y' and substitute those formulas for y' and y into the equation. As Defennder said, you can't actually "find" y in terms of elementary functions and you don't need to. Since you have a product of functions, use the product rule. Of course, the derivative of e^{t^2} is, by the chain rule, 2te^{t^2} and the derivative of the integral is, by the fundamental theorem of calculus, e^{-t^2}.

 

[MustangGt94]

Awesome that works out well, thanks a lot guys!