Answer
$y = \frac{1}{3t^3}~(1+t^2)^{3/2}+\frac{C}{t^3}$, $~~~~t\gt 0$
Work Step by Step
A first-order linear differential equation is one that can be put into the form:
$\frac{dy}{dx} + P(x)~y = Q(x)$
We can consider the given equation:
$t^2\frac{dy}{dt}+3ty = \sqrt{1+t^2}$, $~~~~t \gt 0$
$\frac{dy}{dt}+\frac{3}{t}y = \frac{\sqrt{1+t^2}}{t^2}$, $~~~~t \gt 0$
Let $P(t) = \frac{3}{t}$
Let $Q(t) = \frac{\sqrt{1+t^2}}{t^2}$
We can find the integrating factor:
$I(x) = e^{\int \frac{3}{t} ~dt} = e^{3ln~t} = e^{ln~t^3} = t^3$
We can multiply by the integrating factor:
$t^3~y'+3t^2~y = t~\sqrt{1+t^2}$
$\frac{d}{dt}(t^3~y) = t~\sqrt{1+t^2}$
We can integrate both sides of the equation:
$t^3~y = \int t~\sqrt{1+t^2}~dt$
Let $u = 1+t^2$
$\frac{du}{dt} = 2t$
$dt = \frac{du}{2t}$
$t^3~y = \int t~\sqrt{u}~\frac{du}{2t}$
$t^3~y = \int \frac{1}{2}~\sqrt{u}~du$
$t^3~y = \frac{1}{3}~u^{3/2}+C$
$t^3~y = \frac{1}{3}~(1+t^2)^{3/2}+C$
$y = \frac{1}{3t^3}~(1+t^2)^{3/2}+\frac{C}{t^3}$, $~~~~t\gt 0$
