Question

If $\int\limits_a^x t y(t) d t=x^2+y(x)$ 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
$x y(x)=2 x-y^{\prime}(x)$
hence
$\frac{d y}{d x}-x y=-2 x \left(y^{\prime}(x)=\frac{d y}{d x} ; y(x)=y\right) $
$I . F=e^{\int-x d x}=e^{\frac{-x^2}{2}} $
$y e^{\frac{-x^2}{2}}=\int-2 x e^{\frac{-x^2}{2}} d x ; e^{-\frac{x^2}{2}}=t \Rightarrow -x e^{-\frac{x^2}{2}} d x=d t ; I=\int 2 d t$
$y e^{\frac{-x^2}{2}}=2 e^{\frac{-x^2}{2}}+c$
$y=2+c e^{\frac{x^2}{2}}$
if $x=a \Rightarrow a^2+y=0 \Rightarrow y=-a^2 $ (from the given equation) hence $-a^2=2+c e^{\frac{a^2}{2}} ; c e^{\frac{a^2}{2}}=-\left(2+a^2\right) ; c=-\left(2+a^2\right) e^{-\frac{a^2}{2}} ; y=2-\left(2+a^2\right) e^{\frac{x^2-a^2}{2}}$