Finding f(t) given an integrating factor...

Question

The below differential equation has an integration factor $\mu(t) = t$.

$$ f(t)\frac{dy}{dt} + t^2 + y = 0 $$

My task is to find f(t). I know that multiplying $\mu(t)$ with the diff. equation gives us an exact one, like so...

$$ tf(t)\frac{dy}{dt} + (t^3 \ + \ ty) = 0 $$

...and by definition of exact equations, $tf'(t) + f(t) = t$. How do I proceed on from there? Thank you for helping out!

Answer 1

Rewriting your equation gives;

$\frac{dy}{dt} + \frac{y}{f(t)} =\frac{-t^2}{f(t)}$

the integrating factor is given by ;

$IF = e^{\int\frac1{f(t)}\,dt}= t$

$\implies\ln(t) = \int\frac1{f(t)}\,dt$

$\implies \frac{d{\ln(t)}}{dt} =\frac1{f(t)} $

$\therefore f(t) = t$

Answer 2

You have that $$f't+f=t $$ $$\implies (ft)'=t \implies \int dft=\int tdt \implies f=\frac t2 +\frac Ct$$

Answer 3

The differential equation can be put into the following form: \begin{align} f(t) y' + y &= -t^2 \\ \implies y' + \frac{1}{f(t)}y & = \frac{-t^2}{f(t)} \\ \end{align}

Notice that the integration factor for the above equation is $e^{\int \frac{1}{f(t)}\mathrm{d}t}$. This is because multiplication of the equation with this factor turns it into an equation which is exact:

$$ \hspace{2.2 cm} (y e^{\int \frac{1}{f(t)}\mathrm{d}t})' = \frac{-t^2}{f(t)} e^{\int \frac{1}{f(t)}\mathrm{d}t}$$

Now comparing the multiplication factors, $e^{\int \frac{1}{f(t)}\mathrm{d}t} = t$. This can be solved for $f$.