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

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- 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 . An "integrating factor" is a function u(t) such that mutiplying the entire equation by it:

will make the left side an "exact" derivative: .

By the product rule, and we want that equal to . That means we must have which is a separable equation for u: . Integrating both sides, (since we are only looking for a single solution we can ignore the constant of integration) or .

Multiplying the original equation by that gives .

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

Integrate both sides of that (The left side is easy. Use the substitution on the right.) and solve for y. - Mar 24th 2009, 02:05 PMKrizalid
- Mar 25th 2009, 08:26 AMJhevon