Question
If $$y - y\left( x \right)$$ is the solution of the differential equation, $$x\frac{{dy}}{{dx}} + 2y = \,{x^2}$$ satisfying $$y\left( a \right) = 1,$$ then $$y\left( {\frac{1}{2}} \right)$$ is equal to:
A.
$$\frac{7}{{64}}$$
B.
$$\frac{1}{{4}}$$
C.
$$\frac{49}{{16}}$$
D.
$$\frac{13}{{16}}$$
Answer :
$$\frac{49}{{16}}$$
Solution :
$$\eqalign{
& {\text{Since, }}x\frac{{dy}}{{dx}} + 2y = \,{x^2} \cr
& \Rightarrow \frac{{dy}}{{dx}} + \frac{2}{x}y = x \cr
& {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\frac{2}{x}\,dx} }} = {e^{2\,\ln \,x}} = {e^{\ln \,{x^2}}} = {x^2} \cr} $$
Solution of differential equation is:
$$\eqalign{
& y\,.\,\% \,{x^2} = \int {x.{x^2}dx} \cr
& y\,.\,\% \,{x^2} = \frac{{{x^4}}}{4} + C.....\left( 1 \right) \cr
& \because y\left( a \right) = 1 \cr
& \therefore C = \frac{3}{4} \cr} $$
Then, from equation (1)
$$\eqalign{
& y.\,\% \,{x^2} = \frac{{{x^4}}}{4} + \frac{3}{4} \cr
& \therefore y = \frac{{{x^2}}}{4} + \frac{3}{{4{x^2}}} \cr
& \therefore y\left( {\frac{1}{2}} \right) = \frac{1}{{16}} + 3 = \frac{{49}}{{16}} \cr} $$