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

Printable View

- Mar 23rd 2009, 11:11 PMsazzy3Help with Integrating Factor
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 - Mar 23rd 2009, 11:33 PMJhevon
- Mar 23rd 2009, 11:39 PMsazzy3I have to solve that equation using integrating factor method
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 - Mar 24th 2009, 12:37 AMJhevon
- Mar 24th 2009, 01:06 AMsazzy3continued
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? - Mar 24th 2009, 03:54 AMHallsofIvy
Write the equation as $\displaystyle \frac{dy}{dt}- t^2 y= -t^2$. An "integrating factor" is a function u(t) such that mutiplying the entire equation by it:

$\displaystyle u(t)\frac{dy}{dt}- t^2u(t)y= -t^2u(t)$ will make the left side an "exact" derivative: $\displaystyle \frac{d(u(t)y)}{dt}$.

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

Multiplying the original equation by that gives $\displaystyle e^{-\frac{t^3}{3}}\frac{dy}{dt}- t^2e^{-\frac{t^3}{3}}y= -t^2e^{-\frac{t^3}{3}}$.

Of course, the left side of that is a single derivative:

$\displaystyle \frac{d(e^{-\frac{t^3}{3}})y}{dt}= -t^2e^{-\frac{t^3}{3}}$

Integrate both sides of that (The left side is easy. Use the substitution $\displaystyle v= \frac{t^3}{3}$ on the right.) and solve for y. - Mar 24th 2009, 02:05 PMKrizalid
- Mar 25th 2009, 08:26 AMJhevon