Hi

I am having troubles with proving that y=3t+t^2 is one of the solutions of ty'-y=t^2. Can someone show me? I already solved the equation without the book's help but I don't know what to do with it

Printable View

- Jan 27th 2008, 05:30 PMMathIsRearryHardDifferential Equations Help
Hi

I am having troubles with proving that y=3t+t^2 is one of the solutions of ty'-y=t^2. Can someone show me? I already solved the equation without the book's help but I don't know what to do with it - Jan 27th 2008, 05:40 PMJhevon
- Jan 27th 2008, 05:40 PMtopsquark
I'm confused about something. You say you solved the differential equation, but you don't know if the given solution is one of them? I would think you'd simply be able to compare them.

Anyway, proving the given solution solves the differential equation is easy: Simply stick the solution into the equation:

and see if it is an identity.

So

as desired.

-Dan - Jan 27th 2008, 05:50 PMMathIsRearryHard
Thanks guys. I was overthinking it too much.

I solved the equation and I got this wierd answer (?):

y= -t-1 +Ce^t

How is it possible to substitute with this if we don't know what "C" is? I just wanna know - Jan 27th 2008, 05:55 PMJhevon
- Jan 27th 2008, 06:11 PMMathIsRearryHard
ty'-y+t^2

Ok, this is my first time dealing with equations like these so I might be wrong. Here's how I did it. (A little extra info about my background: I've only been exposed to these type of problems for 2 1/2 days)

The integrating factor here is e^-t since p(t)= -1 and integrating you would get -t.

So,

(ye^-t)'= te^-t

u=t du= dx dv=e^-t dx

Integrating both sides, I finally get

y= -t- 1 +Ce^t

Please tell me if this is wrong when you see it - Jan 27th 2008, 06:22 PMJhevon
to apply the integrating factor method, it's best, if not required, to have the coefficient of y' to be 1. so we would divide through by t before doing anything.

you should have noticed what you had does not work. since if we multiply through by we get:

clearly the left hand side is not the derivative of something using the product rule. the would ensure that the derivative has three terms when fully expanded.

however, if we divide by t first, we have

then , multiplying through we get:

and now the left side is the derivative given by the product rule of

and we're in business - Jan 27th 2008, 06:29 PMMathIsRearryHard
I understand now how you got that but I'm a little shaky with these IF's because I tend to look at them by using the formula e^integral p(t) dx. I did divide by t on both sides before I multiplied the IF on both sides.

Thanks for the help. That is certainly something I've never seen before. - Jan 27th 2008, 06:36 PMJhevon
i'm not sure what's your hangup. just put it in your mind that the ideal form to have is , here is the integrating factor, and you are to multiply through the whole equation by it.

see post #21 here