Ascription to an initial value problem.
Identify the differentiable function $\mu :(0,\infty)\rightarrow \mathbb{R} $, which satisfies
$$\mu (t)=t^2\int _1^t\frac{1-\mu '(\tau )}{\tau ^3}d\tau $$
for all $t > 0$.
I'm having difficulty understanding the approach for solving this particular problem and would be very grateful if someone could provide an impulse to guide my thinking.
On
$$\frac {\mu (t)}{t^2} = \int _1^t \frac{1-\mu '(\tau )}{\tau ^3}d\tau $$ $$\frac{d}{dt}\left (\frac {\mu (t)}{t^2}\right ) =\frac{d}{dt}\left(\int _1^t \frac{1-\mu '(\tau )}{\tau ^3}d\tau \right )=\left [\gamma (\tau )=\frac{1-\mu '(\tau)}{\tau ^3}\right ]=\frac{d}{dt}\left (\Gamma (\tau) \bigg| _{\tau =1}^{\tau =t} \right )$$ $$\frac {t^2\cdot \mu '(t)-2t\cdot \mu (t)}{t^4}=\frac {d}{dt}\big ( (\Gamma (t)-\Gamma (1)\big ) =\gamma (t)=\frac {1-\mu' (t)}{t^3}$$
$$\frac {t^2\cdot \mu '(t)+2t\cdot \mu (t)}{t^4}=\frac {1-\mu '(t)}{t^3}$$ $$\mu '(t)-\frac {2}{t+1}\cdot \mu (t)=\frac{1}{t+1}$$
Homogeneous solution: $$\mu '(t)-\frac {2}{t+1}\cdot \mu (t)=0$$ $$\int \frac {\mu '(t)}{\mu (t)}dt-2\int \frac{1}{t+1}dt=C$$ $$\mu (t)=\chi \cdot (t+1)^2$$
Inhomogeneous solution: $$\chi '(t)(t+1)^2+\chi(t)\cdot 2\cdot (t+1)-\frac{2}{t+1}\cdot \chi (t)\cdot(t+1)^2=\frac{1}{t+1}$$ $$\chi (t)=\int \frac{1}{(t+1)^3}dt$$ $$\chi (t)=-\frac{1}{2\cdot (t+1)^2}+\theta$$
Initial value problem: $$\mu (1)=0 \Rightarrow \theta = \frac{1}{2}$$ $$\mu (t)=\left (-\frac{1}{2\cdot (t+1)^2}+\frac{1}{2}\right )\cdot (t+1)^2$$
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Thank you all for your help. Here is my solution. If anyone notices anything wrong at a glance, I would appreciate corrections.
This is equivalent to $\frac{\mu(t)}{t^2}=\int_1^t\frac{1-\mu'(\tau)}{\tau^3}d\tau$ and by differentiation you get $\frac{\mu'(t)}{t^2}-\frac{2\mu(t)}{t^3}=\frac{1-\mu'(t)}{t^3}$ so a necessary condition for $\mu$ is to be solution of the following classical differential equation : $$(t+1)\mu'(t)-2\mu(t)=1$$ with condition $\mu(1)=0$.