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

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- Jan 27th 2008, 04:30 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, 04:40 PMJhevon
- Jan 27th 2008, 04:40 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:

$\displaystyle ty^{\prime} - y = t^2$

and see if it is an identity.

$\displaystyle y = 3t + t^2$

$\displaystyle y^{\prime} = 3 + 2t$

So

$\displaystyle t(3 + 2t) - (3t + t^2)$

$\displaystyle = 3t + 2t^2 - 3t - t^2$

$\displaystyle = t^2$

as desired.

-Dan - Jan 27th 2008, 04:50 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, 04:55 PMJhevon
- Jan 27th 2008, 05:11 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, 05:22 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 $\displaystyle e^{-t}$ we get:

$\displaystyle te^{-t}y' - e^{-t}y = t^2e^{-t}$

clearly the left hand side is not the derivative of something using the product rule. the $\displaystyle te^{-t}$ would ensure that the derivative $\displaystyle te^{-t}y$ has three terms when fully expanded.

however, if we divide by t first, we have $\displaystyle y' - \frac 1ty = t$

then $\displaystyle \mu (t) = exp \left( - \int \frac 1t~dt \right) = \frac 1t$, multiplying through we get:

$\displaystyle \frac 1ty' - \frac 1{t^2}y = 1$

and now the left side is the derivative given by the product rule of $\displaystyle \frac 1ty$

and we're in business - Jan 27th 2008, 05:29 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, 05:36 PMJhevon
i'm not sure what's your hangup. just put it in your mind that the ideal form to have is $\displaystyle y' + p(t)y = g(t)$, here $\displaystyle exp \left( \int p(t)~dt \right)$ is the integrating factor, and you are to multiply through the whole equation by it.

see post #21 here