• # question_answer The solution of the equation$x\int\limits_{0}^{x}{y(t)dt=(x+1)\int_{0}^{x}{ty(t)dt,x>0}}$ is A) $y=\frac{c}{{{x}^{3}}}{{e}^{{{x}^{3}}}}$ B) $y=c{{x}^{3}}{{e}^{-{{x}^{3}}}}$ C) $\frac{c}{{{x}^{3}}}{{e}^{-x}}$ D) None of these

 [d] The equation is $x\int_{0}^{x}{y(t)dt=(x+1)\int_{0}^{x}{ty(t)dt,}}$               ? (i) Differentiating both the sides with respect to x, we get $xy(x)+\int_{0}^{x}{y(t)dt=(x+1)xy(x)+\int_{0}^{x}{ty(t)dt}}$ $\Rightarrow \int_{0}^{x}{y(t)dt={{x}^{2}}y(x)+\int_{0}^{x}{ty(t)dt}}$                    ? (ii) Differentiating again with respect to x, we get $y(x)=2xy(x)+{{x}^{2}}y'(x)+xy(x)$ $\Rightarrow \,\,\,{{x}^{2}}\frac{dy}{dx}=(1-3x)y$       $[writing\,\,y(x)=y]$ $\Rightarrow \frac{dy}{y}=\left( \frac{1-3x}{{{x}^{2}}} \right)dx=\left( \frac{1}{{{x}^{2}}}-\frac{3}{x} \right)dx$ Integrating we get, $\log y=-\frac{1}{x}-3\log x+a,a$ is constant $\Rightarrow \log y+3\log x=a-\frac{1}{x}$ $\Rightarrow \log (y{{x}^{3}})=a-\frac{1}{x}\Rightarrow y{{x}^{3}}={{e}^{a-\frac{1}{x}}}=c.{{e}^{-\frac{1}{x}}}$ where $c={{e}^{a}}\therefore y=\frac{c}{{{x}^{3}}}{{e}^{-\frac{1}{x}}}$