Question

The solution of the differential equation $\frac{dy}{dx}+\frac{1}{x}=\frac{e^y}{x^2},\left(\forall x>0\right)$ is $\lambda x{e}^{-y}=1-2c{x}^2$ (where $c$ is an arbitrary constant). Then, the value of $\lambda$ is equal to

• A. $2$
• B. $4$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

The given equation is $e^{-y}\frac{dy}{dx}+e^{-y}\cdot \frac{1}{x}=\frac{1}{x^2}$
Let, $-e^{-y}=t\Rightarrow e^{-y}\frac{dy}{dx}=\frac{dt}{dx}$
Thus we have,
$\frac{dt}{dx}-t\frac{1}{x}=\frac{1}{x^2}$
Integrating factor $=e^{-\int \frac{1}{x}dx}=e^{-lnx}=\frac{1}{x}$
Thus, the solution is
$t\cdot \left(\frac{1}{x}\right)=\int \frac{1}{x^2}\cdot \frac{1}{x}dx$
$\Rightarrow \frac{-e^{-y}}{x}=\frac{x^{-2}}{-2}+c$
$\Rightarrow \frac{-e^{-y}}{x}=\frac{-1}{2x^2}+c$
$\Rightarrow -2x{e}^{-y}=-1+2c{x}^2$ or $2x{e}^{-y}=1-2c{x}^2$
Hence, $\lambda =2$