JEE Main 2020 — Mathematics Calculus

Given $x\frac{dy}{dx}-y={x}^{2}(\mathrm{xcos}+\mathrm{sinx})$xdxdy​−y=x2(xcos+sinx)

$\Rightarrow$⇒$\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{x}y=x(\mathrm{xcosx}+\mathrm{sinx})$dxdy​−x1​y=x(xcosx+sinx)

$\therefore$∴ I.F $={e}^{-l\mathrm{nx}}=\frac{1}{x}$=e−lnx=x1​

$\therefore$∴ Solution is $\displaystyle y.\frac{1}{x}=\int \frac{1}{x}.x(x\mathrm{cos}x+\mathrm{sin}x)dx$y.x1​=∫x1​.x(xcosx+sinx)dx

$$\frac{y}{x}=\int (x\mathrm{cos}x+\mathrm{sin}x)\mathrm{dx}$$xy​=∫(xcosx+sinx)dx

$$\frac{y}{x}=x\mathrm{sin}x+C$$xy​=xsinx+C

$$\because y(\pi )=\pi \Rightarrow C=1$$∵y(π)=π⇒C=1

$$y={x}^{2}\mathrm{sin}x+x$$y=x2sinx+x

$$\frac{dy}{dx}={x}^{2}\mathrm{cos}x+2x\mathrm{sin}x+1$$dxdy​=x2cosx+2xsinx+1

$$\frac{{d}^{2}y}{d{x}^{2}}=-{x}^{2}\mathrm{sin}x+2x\mathrm{cos}x+2\mathrm{sin}x+2x\mathrm{cos}x$$dx2d2y​=−x2sinx+2xcosx+2sinx+2xcosx

$$=-{x}^{2}\mathrm{sin}x+4x\mathrm{cos}x+2\mathrm{sin}x$$=−x2sinx+4xcosx+2sinx

$\therefore$∴ ${y}^{''}(\frac{\pi }{2})+y(\frac{\pi }{2})=(-\frac{{\pi }^{2}}{4}+0+2)+(\frac{{\pi }^{2}}{4}+\frac{\pi }{2})$y′′(2π​)+y(2π​)=(−4π2​+0+2)+(4π2​+2π​)

$$=-\frac{{\pi }^{2}}{4}+2+\frac{{\pi }^{2}}{4}+\frac{\pi }{2}$$=−4π2​+2+4π2​+2π​

$=\frac{\pi }{2}+2$=2π​+2.