Differential Equations 2 Question 6
6. The solution of the differential equation $x \frac{d y}{d x}+2 y=x^{2}(x \neq 0)$ with $y(1)=1$, is $\quad$ (2019 Main, 9 April I)
(a) $y=\frac{x^{2}}{4}+\frac{3}{4 x^{2}}$
(b) $y=\frac{x^{3}}{5}+\frac{1}{5 x^{2}}$
(c) $y=\frac{3}{4} x^{2}+\frac{1}{4 x^{2}}$
(d) $y=\frac{4}{5} x^{3}+\frac{1}{5 x^{2}}$
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Answer:
Correct Answer: 6. (a)
Solution:
- Given differential equation is
$$ \begin{aligned} & x \frac{d y}{d x}+2 y=x^{2},(x \neq 0) \\ \Rightarrow \quad \frac{d y}{d x}+\frac{2}{x} y & =x, \end{aligned} $$
which is a linear differential equation of the form
$$ \frac{d y}{d x}+P y=Q $$
Here, $P=\frac{2}{x}$ and $Q=x$
$$ \therefore \quad IF=e^{\int \frac{2}{x} d x}=e^{2 \log x}=x^{2} $$
Since, solution of the given differential equation is
$$ \begin{aligned} & y \times IF=\int(Q \times IF) d x+C \\ \therefore & y\left(x^{2}\right)=\int\left(x \times x^{2}\right) d x+C \Rightarrow y x^{2}=\frac{x^{4}}{4}+C \\ \because & y(1)=1, \text { so } 1=\frac{1}{4}+C \Rightarrow C=\frac{3}{4} \\ \therefore & \quad y x^{2}=\frac{x^{4}}{4}+\frac{3}{4} \Rightarrow y=\frac{x^{2}}{4}+\frac{3}{4 x^{2}} \end{aligned} $$
