JEE Main 2016 — Mathematics Calculus

$$x\int _{1}^{x}y(t)dt=x\int _{1}^{x}ty(t)dt+ \int _{1}^{x}ty(t)dt$$x∫1x​y(t)dt=x∫1x​ty(t)dt+∫1x​ty(t)dt

Differentiate w.r.t. $x$x

$$\int _{1}^{x}y(t)dt+xy(x)=\int _{1}^{x}ty(t)dt+x[xy(x)]+xy(x)$$∫1x​y(t)dt+xy(x)=∫1x​ty(t)dt+x[xy(x)]+xy(x)

$$\int _{1}^{x}y(t)dt=\int _{1}^{x}ty(t)dt+{x}^{2}y(x)$$∫1x​y(t)dt=∫1x​ty(t)dt+x2y(x)

Differentiate again w.r.t. $x$x

$$y(x)=xy(x)+2x y(x)+{x}^{2}{y}^{'}(x)$$y(x)=xy(x)+2xy(x)+x2y′(x)

$$(1-3x)y(x)={x}^{2}{y}^{'}(x)$$(1−3x)y(x)=x2y′(x)

$$\frac{{y}^{'}(x)}{y(x)}=\frac{1-3x}{{x}^{2}}$$y(x)y′(x)​=x21−3x​

$$\frac{1}{y}\frac{dy}{dx}=\frac{1-3x}{{x}^{2}}$$y1​dxdy​=x21−3x​

Integrating on both sides

$$\Rightarrow \mathrm{ln}y=-\frac{1}{x}-3\mathrm{ln}x+C$$⇒lny=−x1​−3lnx+C

$$\Rightarrow \mathrm{ln}(y{x}^{3})=-\frac{1}{x}+C$$⇒ln(yx3)=−x1​+C

$$\Rightarrow y{x}^{3}= {e}^{-\frac{1}{x}+C}$$⇒yx3=e−x1​+C

$$\Rightarrow y=\frac{{e}^{-\frac{1}{x}+C}}{{x}^{3}}$$⇒y=x3e−x1​+C​

$$\Rightarrow y=\frac{C}{{x}^{3}}{e}^{-\frac{1}{x}}$$⇒y=x3C​e−x1​