Question

If $y=100+f\left(x\right)$ is the particular solution of differential equation $\frac{dy}{dx}=100-y$ and $y\left(0\right)=50,$ then the value of $f\left(1\right)$ is

• A. $50e$
  
• B. $-50e$
• C. $-\frac{50}{e}$
• D. $\frac{50}{e}$

Answer 1

The correct option is C $-\frac{50}{e}$

$\int \frac{dy}{100-y}=\int dx$

$\Rightarrow -\ln |100-y|=x+C_1,C_1$ is a constant.
$\Rightarrow \ln |100-y|=-x+C(C=-C_1)$
where $C$ is a constant.
If $x=0,$ then $y=50$ hence, $C=\ln 50$
$x=\ln 50-\ln |100-y|$
$\Rightarrow \ln \left|\frac{50}{100-y}\right|=x$
$\Rightarrow \frac{50}{100-y}=e^x$
$\Rightarrow 100-y=50e^{-x}$
$\Rightarrow y=100-50e^{-x}$
$\Rightarrow f\left(x\right)=-50e^{-x}$
$\Rightarrow f\left(1\right)=-\frac{50}{e}$