Question

For $xϵR,x\ne 0$, if $y\left(x\right)$ is a differentiable function such that $x{\int }_1^{x}y\left(t\right)dt=(x+1){\int }_1^{x}ty\left(t\right)dt$, then $y\left(x\right)$ equals:
(Where C is a constant)

• A. $C{x}^3{e}^{\frac{1}{x}}$
• B. $\frac{C}{x^2}{e}^{-\frac{1}{x}}$
• C. $\frac{C}{x}{e}^{-\frac{1}{x}}$
• D. $\frac{C}{x^3}{e}^{-\frac{1}{x}}$

Answer 1

Answer: (D)

$\frac{C}{x^3}{e}^{-\frac{1}{x}}$

$x{\int }_1^{x}y\left(t\right)dt=x{\int }_1^{x}ty\left(t\right)dt+{\int }_1^{x}ty\left(t\right)dt$
differentiate w.r. to x.
${\int }_1^{x}y\left(t\right)dt+x\left[y\right(x)-y(1\left)\right]={\int }_1^{x}ty\left(t\right)dt+x\left[xy\right(x)-y(1\left)\right]+xy\left(x\right)-y\left(1\right)$
${\int }_1^{x}y\left(t\right)dt={\int }_1^{x}ty\left(t\right)dt+x^{2}y\left(x\right)-y\left(1\right)$
diff. again w.r to x
$y\left(x\right)-y\left(1\right)=xy\left(x\right)-y\left(1\right)+2xy\left(x\right)+x^2{y}^{{}^{\prime }}\left(x\right)$
$(1-3x)y\left(x\right)=x^2{y}^{{}^{\prime }}\left(x\right)$
$\frac{y^{{}^{\prime }}\left(x\right)}{y\left(x\right)}=\frac{1-3x}{x^2}$
$\frac{1}{y}\frac{dy}{dx}=\frac{1-3x}{x^2}\Rightarrow lny=-\frac{1}{x}-3lnx$
$ln\left(y{x}^3\right)=-\frac{1}{x}+lnc$
$y{x}^3=c{e}^{-\frac{1}{x}}$
$y=c\frac{e^{-\frac{1}{x}}}{x^3}$ or $y=\frac{c{e}^{-\frac{1}{x}}}{x^3}$