Question

If $\underset{a}{\overset{x}{\int }}ty(t)dt=x^2+y(x)$ then $y$ as a function of $x$ is

• A. $y=2-\left(2+a^2\right)e^{\frac{x^2-a^2}{2}}$
• B. $y=1-\left(2+a^2\right)e^{\frac{x^2-a^2}{2}}$
• C. $y=2-\left(1+a^2\right)e^{\frac{x^2-a^2}{2}}$
• D. none

Answer 1

Answer: (A)

diff. both sides
$xy(x)=2x-y^{\prime }(x)$
hence
$\frac{dy}{dx}-xy=-2x\left(y^{\prime }(x)=\frac{dy}{dx};y(x)=y\right)$
$I.F=e^{\int -xdx}=e^{\frac{-x^2}{2}}$
$y{e}^{\frac{-x^2}{2}}=\int -2x{e}^{\frac{-x^2}{2}}dx;e^{-\frac{x^2}{2}}=t\Rightarrow -x{e}^{-\frac{x^2}{2}}dx=dt;I=\int 2dt$
$y{e}^{\frac{-x^2}{2}}=2e^{\frac{-x^2}{2}}+c$
$y=2+c{e}^{\frac{x^2}{2}}$
if $x=a\Rightarrow a^2+y=0\Rightarrow y=-a^2$ (from the given equation) hence $-a^2=2+c{e}^{\frac{a^2}{2}};c{e}^{\frac{a^2}{2}}=-\left(2+a^2\right);c=-\left(2+a^2\right)e^{-\frac{a^2}{2}};y=2-\left(2+a^2\right)e^{\frac{x^2-a^2}{2}}$