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)

$\underset{a}{\overset{x}{\int }}y(t)dt=x^2+y(x)$
$\Rightarrow xy=2x+\frac{dy}{dx}\Rightarrow x(y-2)=\frac{dy}{dx}$
$\Rightarrow \int xdx=\int \frac{dy}{y-2}\Rightarrow \frac{x^2}{2}=\ell n|y-2|+\ell nc$
$\Rightarrow e^{\frac{x^2}{2}}=c(y-2)$
$atx=ay=-a^2$
$\therefore e^{\frac{a^2}{2}}=c\left(-a^2-2\right)\Rightarrow c=-\frac{e^{\frac{a^2}{2}}}{\left(a^2+2\right)}$
$\therefore e^{\frac{x^2}{2}}=-\frac{e^{\frac{a^2}{2}}}{\left(a^2+2\right)}(y-2)$
$\Rightarrow -y+2=\left(a^2+2\right)e^{\frac{x^2-a^2}{2}}$
$\Rightarrow y=2-\left(2+a^2\right)e^{\frac{x^2-a^2}{2}}$