Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

NTA AbhyasNTA Abhyas 2022

Solution:

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