Q120 / 160 · TS EAMCET 2016 · Mathematics
The solution of the differential equation $y^{\prime}=\frac{1}{e^{-y}-x}$ is
- Correct answerA$x=e^{-y}(y+C)$
- B$y+e^{-y}=x+C$
- C$x=e^y(y+C)$
- D$x+y=e^{-y}+C$
Solution
Given differential equation is $y^{\prime}=\frac{1}{e^y-x}$
$$
\frac{d y}{d x}=\frac{1}{e^y-x}
$$
We can write this equation in the form
$$
\begin{aligned}
\frac{d x}{d y} & =e^y-x \\
\frac{d x}{d y}+x & =e^y
\end{aligned}
$$
Compare Eq. (i) by $\frac{d x}{d y}+P x=Q$, we get
$$
\begin{aligned}
P & =1, Q=e^y \\
\therefore \quad \text { If }=e^{\int P d y} & =e^{\int 10 y}=e^y
\end{aligned}
$$
Solution of given differential equation is
$$
\begin{aligned}
x \cdot e^y & =\int e^y \cdot e^y d y+c \\
x \cdot e^y & =\int 1 \cdot d y+c
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & x e^y=y+C \\
\therefore & x=e^y(y+c)
\end{array}
$$
Hence, the correct answer is option: 'A'.