# Differential Equations Help

• January 27th 2008, 04:30 PM
MathIsRearryHard
Differential 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
• January 27th 2008, 04:40 PM
Jhevon
Quote:

Originally Posted by MathIsRearryHard
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

it's easier than it seems

just find y' and then plug in y and y' into the left hand side of the differential equation and show that you get the right hand side
• January 27th 2008, 04:40 PM
topsquark
Quote:

Originally Posted by MathIsRearryHard
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

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:
$ty^{\prime} - y = t^2$
and see if it is an identity.

$y = 3t + t^2$

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

So
$t(3 + 2t) - (3t + t^2)$

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

$= t^2$
as desired.

-Dan
• January 27th 2008, 04:50 PM
MathIsRearryHard
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
• January 27th 2008, 04:55 PM
Jhevon
Quote:

Originally Posted by MathIsRearryHard
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

how exactly did you get that solution. i did it and got $y = t^2 + Ct$, which is the correct form.

(we use the integrating factor method here)
• January 27th 2008, 05:11 PM
MathIsRearryHard
Quote:

Originally Posted by Jhevon
how exactly did you get that solution. i did it and got $y = t^2 + Ct$, which is the correct form.

(we use the integrating factor method here)

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
• January 27th 2008, 05:22 PM
Jhevon
Quote:

Originally Posted by MathIsRearryHard
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

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 $e^{-t}$ we get:

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

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

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

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

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

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

• January 27th 2008, 05:29 PM
MathIsRearryHard
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.
• January 27th 2008, 05:36 PM
Jhevon
Quote:

Originally Posted by MathIsRearryHard
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.

i'm not sure what's your hangup. just put it in your mind that the ideal form to have is $y' + p(t)y = g(t)$, here $exp \left( \int p(t)~dt \right)$ is the integrating factor, and you are to multiply through the whole equation by it.

see post #21 here