Question

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

• 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 $-\frac{1}{2}$

Here, $\frac{dy}{dt}-\left(\frac{t}{1+t}\right)y=\frac{1}{1+t}$ and $y\left(0\right)=-1$
which represents linear differential equation of first order.
$\text{IF}=e^{-\int \left(\frac{t}{1+t}\right)dt}=e^{-\int \frac{(t+1+1)}{1+t}=dt}$
$=e^{-t+\log (1+t)}$
$=e^{-t}(1+t)$
Required solution is $y\left(\text{IF}\right)=[\int Q\cdot (\text{IF}\left)dt\right]+C$
$y{e}^{-t}(1+t)=\int \frac{1}{1+t}{e}^{-t}(1+t)dt+C$
$=\int e^{-t}dt+C=e^{-t}+C$
At $t=0,y=-1$
Therefore, $-1(1+0)=-e^0+C$
$\Rightarrow C=0$
Therefore, $y{e}^{-t}(1+t)=-e^{-t}$
At $t=1$
$y{e}^{-1}(1+1)=-e^{-1}$
$\Rightarrow y=-\frac{1}{2}$.