Help with Integrating Factor

**sazzy3** · March 23rd 2009, 11:11 PM

Can anyone help me solve this equation using the method differentiation with integrating factor:

dy/dt=(y-1)t^2, where inital condition is y(1)=2

**Jhevon** · March 23rd 2009, 11:33 PM

Originally Posted by sazzy3

Can anyone help me solve this equation using the method differentiation with integrating factor:

dy/dt=(y-1)t^2, where inital condition is y(1)=2

this has nothing to do with integrating factors. the equation is separable

it can be written as:

$\frac {dy}{y - 1} = t^2~dt$

now integrate both sides and continue

**sazzy3** · March 23rd 2009, 11:39 PM

I have to solve this equation using the method of integrating factor.
And i have no idea how
i know i have to change the equation to the form to linear
A1(t)dy/dt + A0(t)y=f(t)
then write it as: dy/dt + p(t)y=q(t)
where p(t)=A0(t)/A1(t) and q(t)=f(t)/A1(t)

Then solve by finding an integrating factor etc

**Jhevon** · March 24th 2009, 12:37 AM

Originally Posted by sazzy3

I have to solve this equation using the method of integrating factor.
And i have no idea how
i know i have to change the equation to the form to linear
A1(t)dy/dt + A0(t)y=f(t)
then write it as: dy/dt + p(t)y=q(t)
where p(t)=A0(t)/A1(t) and q(t)=f(t)/A1(t)

Then solve by finding an integrating factor etc

fine, it is overkill though

you have $y' = (y - 1)t^2 = yt^2 - t^2$

$\Rightarrow y' - t^2y = -t^2$

now you have your $p(t) = -t^2$ . the integrating factor is given by $e^{\int p(t)}$

now see post #2 here

you may also want to see post #21 here

**sazzy3** · March 24th 2009, 01:06 AM

So i get from
having alpha(t)= e^ integral from t to t0 -t^2 dt

then
alpha(t)= exp [integral from t to t0 -t^2 dt]

then exp -t^3/3=-t^2
then

-t^2dy/dt -t^2y=-t^2 x -t^2

is this correct so far i am now lost as to where to go from here?

**HallsofIvy** · March 24th 2009, 03:54 AM

Originally Posted by sazzy3

Can anyone help me solve this equation using the method differentiation with integrating factor:

dy/dt=(y-1)t^2, where inital condition is y(1)=2

Write the equation as $\frac{dy}{dt}- t^2 y= -t^2$ . An "integrating factor" is a function u(t) such that mutiplying the entire equation by it:
$u(t)\frac{dy}{dt}- t^2u(t)y= -t^2u(t)$ will make the left side an "exact" derivative: $\frac{d(u(t)y)}{dt}$ .

By the product rule, $\frac{d(u(t)y)}{dt}= u(t)\frac{dy}{dt}+ \frac{du}{dt}y$ and we want that equal to $u(t)\frac{dy}{dt}- t^2u(t)y$ . That means we must have $\frac{du}{dt}= -t^2u$ which is a separable equation for u: $\frac{du}{u}= -t^2dt$ . Integrating both sides, $ln(u)= -\frac{1}{3}t^3$ (since we are only looking for a single solution we can ignore the constant of integration) or $u(t)= e^{-\frac{t^3}{3}}$ .

Multiplying the original equation by that gives $e^{-\frac{t^3}{3}}\frac{dy}{dt}- t^2e^{-\frac{t^3}{3}}y= -t^2e^{-\frac{t^3}{3}}$ .
Of course, the left side of that is a single derivative:
$\frac{d(e^{-\frac{t^3}{3}})y}{dt}= -t^2e^{-\frac{t^3}{3}}$
Integrate both sides of that (The left side is easy. Use the substitution $v= \frac{t^3}{3}$ on the right.) and solve for y.

**Krizalid** · March 24th 2009, 02:05 PM

Originally Posted by Jhevon

now see post #2 here

you may also want to see post #21 here

And I suggest you to put those links into your signature.

**Jhevon** · March 25th 2009, 08:26 AM

Originally Posted by Krizalid

And I suggest you to put those links into your signature.

Haha, yes. you have been telling me that for quite some time now. I will put the links I refer people to often in my signature. Save me the trouble of searching for the posts every time!

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