Differential Equations Ques 30
- If $y=y(x)$ is the solution of the differential equation, $x \frac{d y}{d x}+2 y=x^{2}$ satisfying $y(1)=1$, then $y \frac{1}{2}$ is equal to
(a) $\frac{13}{16}$
(b) $\frac{1}{4}$
(c) $\frac{49}{16}$
(d) $\frac{7}{64}$
(2019 Main, 9 Jan I)
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Answer:
Correct Answer: 30.(c)
Solution:
Formula:
ELEMENTARY TYPES OF FIRST ORDER \& FIRST DEGREE DIFFERENTIAL EQUATIONS :
- Given differential equation can be rewritten as $\frac{d y}{d x}+\frac{2}{x} \cdot y=x$, which is a linear differential equation of the form $\frac{d y}{d x}+P y=Q$, where $P=\frac{2}{x}$ and $Q=x$.
Now, integrating factor
$\left[\because e^{\log f(x)}=f(x)\right]$
$$ \text { (IF) }=e^{\int \frac{2}{x} d x}=e^{2 \log x}=e^{\log x^{2}}=x^{2} $$
and the solution is given by
$$ \begin{array}{rlrl} & & y(IF) & =\int(Q \times IF) d x+C \\ \Rightarrow \quad y x^{2} & =\int x^{3} d x+C \\ \Rightarrow \quad y x^{2} & =\frac{x^{4}}{4}+C \end{array} $$
Since, it is given that $y=1$ when $x=1$
$\therefore$ From Eq. (i), we get
$$ \begin{array}{rlrl} & & 1 & =\frac{1}{4}+C \Rightarrow C=\frac{3}{4} \\ \therefore & & 4 x^{2} y & =x^{4}+3 \quad \text { [using Eqs. (i) and (ii)] } \\ \Rightarrow \quad y & =\frac{x^{4}+3}{4 x^{2}} \\ \text { Now, } & y & \frac{1}{2} & =\frac{\frac{1}{16}+3}{4 \times \frac{1}{4}}=\frac{49}{16} \end{array} $$
BETA
