Question

If y(t) is a solution of $(1+t)\frac{dy}{dt}-ty=1$ and y(0)=-1, then show that $y\left(1\right)=-\frac{1}{2}.$

Answer 1

Given that, $(1+t)\frac{dy}{dt}-ty=1\Rightarrow \frac{dy}{dt}-\left(\frac{t}{1+t}\right)y=\frac{1}{1+t}$
which is a linear differential equation.
On comparing it with $\frac{dy}{dt}+Py=Q,$ we get
$P=-\left(\frac{t}{1+t}\right),Q=\frac{1}{1+t}$
$IF=e^{-\int \frac{t}{1+t}}dt=e^{-\int (1-\frac{1}{1+t})dt}=e^{-[t-log(1+t\left)\right]}=e^{-t}.e^{log(1+t)}=e^{-t}(1+t)$
The general solution is
$y\left(t\right).\frac{(1+t)}{e^t}=\int \frac{(1+t).e^{-t}}{(1+t)}dt+C\Rightarrow y\left(t\right)=\frac{e^{-t}}{(-1)}.\frac{e^t}{1+t}+C^{\prime },where{C}^{\prime }=\frac{C{e}^{\prime }}{1+t}\Rightarrow y\left(t\right)=-\frac{1}{1+t}+C^{\prime }$
When t=0 and y=-1, then
$-1=-1+C^{\prime }\Rightarrow C^{\prime }=0y\left(t\right)=-\frac{1}{1+t}\Rightarrow y\left(1\right)=-\frac{1}{2}$