Q120 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Mathematics
The general solution of the differential equation $\frac{d x}{d y}+\frac{x}{y}=x^2$ is
- A$\frac{1}{y}=c x-y \log x$
- B$\frac{1}{x}=c y+x \log x$
- Correct answerC$\frac{1}{x}=c y-y \log y$
- D$\frac{1}{y}=c x+y \log x$
Solution
Given differential equation is
$$
\frac{d x}{d y}+\frac{x}{y}=x^2
$$
On dividing by $x^2$, we get
$$
\begin{aligned}
& \frac{1}{x^2} \frac{d x}{d y}+\left(\frac{1}{y}\right)\left(\frac{1}{x}\right)=1 \\
& \text { let } \quad \frac{1}{x}=t \\
& -\frac{1}{x^2} \frac{d x}{d y}=\frac{d t}{d y} \Rightarrow \frac{1}{x^2} \frac{d x}{d y}=-\frac{d t}{d y} \\
& \Rightarrow \quad-\frac{d t}{d y}+\frac{1}{y}(t)=1 \\
& \Rightarrow \quad \frac{d t}{d y}+\left(-\frac{1}{y}\right)(t)=-1 \\
& \text { I. } \mathrm{F}=e^{\int-\frac{1}{y} d y}=e^{-\log y}=\frac{1}{y} \\
&
\end{aligned}
$$
So, solution is
$$
\begin{aligned}
& t \cdot\left(\frac{1}{y}\right)=\int(-1)\left(\frac{1}{y}\right) d y+c \\
& \Rightarrow \quad \frac{t}{y}=-\log y+c \\
& \Rightarrow \quad t=-y \log y+c y \\
& \Rightarrow \quad \frac{1}{x}=c y-y \log y \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'C'.