Q118 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics
The general solution of the differential equation $\left(\frac{1}{x^2}+x\right) \frac{d y}{d x}+3 y=1$ is
- A$y=\frac{1}{x^2}+3 c$
- Correct answerB$(3 y-1) x^3+3 y=c$
- C$\log y-x y=c$
- D$\left(1+x^3\right) y=x^3+c$
Solution
Given differential equation is
$$
\begin{array}{ll}
\Rightarrow & \left(\frac{1}{x^2}+x\right) \frac{d y}{d x}+3 y=1 \\
\Rightarrow & \left(\frac{1+x^3}{x^2}\right) \frac{d y}{d x}+3 y=1 \\
\Rightarrow & \frac{d y}{d x}+\frac{3 x^2}{1+x^3} y=\frac{x^2}{1+x^3}
\end{array}
$$
Which is the linear differential equation of the form $\frac{d y}{d x}+P y=Q$.
Here, $P=\frac{3 x^2}{1+x^3}$ and $Q=\frac{x^2}{1+x^3}$
Now, IF $=\int_e P d x=e^{\int \frac{3 x^2}{1+x^3} d x}=e^{\log \left(1+x^3\right)}$ $=1+x^3$
and solution of the given equation, is
$$
\begin{array}{ll}
y\left(1+x^3\right)= & \int\left(1+x^3\right) \frac{x^2}{\left(1+x^3\right)} d x+k=\int x^2 d x+k \\
\Rightarrow & y\left(1+x^3\right)=\frac{x^3}{3}+k \\
\Rightarrow & 3 y+3 y x^3=x^3+3 k \\
\Rightarrow & (3 y-1) x^3+3 y=3 k \\
\Rightarrow & (3 y-1) x^3+3 y=c \quad \quad[\because 3 k=c]
\end{array}
$$
Hence, the correct answer is option: 'B'.