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
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
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.