If y=y(x) is the solution of the differential equation,

Question:

If $y=y(x)$ is the solution of the differential equation, $x \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=x^{2}$ satisfying $y(1)=1$, then $y\left(\frac{1}{2}\right)$ is equal to:

  1. (1) $\frac{7}{64}$

  2. (2) $\frac{1}{4}$

  3. (3) $\frac{49}{16}$

  4. (4) $\frac{13}{16}$


Correct Option: , 3

Solution:

 Since, $x \frac{d y}{d x}+2 y=x^{2}$

$\Rightarrow \quad \frac{d y}{d x}+\frac{2}{x} y=x$

I.F. $\quad=e^{\int \frac{2}{x} d x}=e^{2 \ln x}=e^{\ln x^{2}}=x^{2}$.

Solution of differential equation is:

$y \cdot x^{2}=\int x \cdot x^{2} d x$

$y \cdot x^{2}=\frac{x^{4}}{4}+C$ ....(1)

$\because \quad y(1)=1$

$\therefore \quad C=\frac{3}{4}$

Then, from equation (1)

$y \cdot x^{2}=\frac{x^{4}}{4}+\frac{3}{4}$

$\therefore \quad y=\frac{x^{2}}{4}+\frac{3}{4 x^{2}}$

$\therefore \quad y\left(\frac{1}{2}\right)=\frac{1}{16}+3=\frac{49}{16}$

 

 

 

 

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