Question

If $y\left(t\right)$ is the solution of the equation $(1+t)\frac{dy}{dt}-ty=1$ and $y\left(0\right)=-1$ then $y\left(1\right)$ is:

• A. $-\frac{1}{2}$
• B. $e+\frac{1}{2}$
• C. $e-\frac{1}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $-\frac{1}{2}$

$(t+1)\frac{dy}{dt}-ty=1\Rightarrow \frac{dy}{dt}-\frac{ty}{(t+1)}=\frac{1}{(t+1)}$
Multiplying both sides by $u$
$u\frac{dy}{dt}-\frac{uty}{(t+1)}=\frac{u1}{(t+1)}$
Substituting $u=e^{-\int -\frac{1}{(t+1)}dt}=e^{-t}(t+1)$
$\frac{(t+1)}{e^t}\frac{dy}{dt}-\frac{ty}{e^t}=e^{-t}$
Substituting $-e^{-t}t=\frac{d}{dt}\left(\frac{t+1}{e^t}\right)$
$\frac{(t+1)}{e^t}\frac{dy}{dt}-\frac{d}{dt}\left(\frac{t+1}{e^t}\right)=e^{-t}$
Using $g\frac{df}{dx}+f\frac{dg}{dx}=\frac{d}{dx}\left(fg\right)$
$\frac{d}{dt}\left(\frac{t+1}{e^t}y\right)=e^{-t}$
Integrating both sides
$\int \frac{d}{dt}\left(\frac{t+1}{e^t}y\right)=\int e^{-t}\Rightarrow \frac{t+1}{e^t}y=-e^{-t}+c\Rightarrow y=\frac{c{e}^t-1}{t+1}y\left(0\right)=-1\Rightarrow c=0y\left(1\right)=-\frac{1}{2}$