JEE Main 2021 — Mathematics Calculus

Let ${e}^{y}=t\Rightarrow {e}^{y}\frac{dy}{dx}=\frac{dt}{dx}$ey=t⇒eydxdy​=dxdt​

So,

$${e}^{y}\frac{dy}{dx}-2{e}^{y}\mathrm{sin}x+\mathrm{sin}x{\mathrm{cos}}^{2}x=0$$eydxdy​−2eysinx+sinxcos2x=0

$$\Rightarrow \frac{dt}{dx}-(2\mathrm{sin}x)t=-\mathrm{sin}x{\mathrm{cos}}^{2}x$$⇒dxdt​−(2sinx)t=−sinxcos2x

So, the integrating factor $\displaystyle I.F.={e}^{\int -2\mathrm{sin}xdx}={e}^{2\mathrm{cos}x}$I.F.=e∫−2sinxdx=e2cosx

So, the solution of differential equation is

$$t\cdot {e}^{2\mathrm{cos}x}=\int {e}^{2\mathrm{cos}x}\cdot (-\mathrm{sin}x{\mathrm{cos}}^{2}x)dx$$t⋅e2cosx=∫e2cosx⋅(−sinxcos2x)dx

$\displaystyle \Rightarrow {e}^{y}\cdot {e}^{2\mathrm{cos}x}=\int {e}^{2z}\cdot {z}^{2}dz$⇒ey⋅e2cosx=∫e2z⋅z2dz (Assume, $z=\mathrm{cos}x\Rightarrow dz=-\mathrm{sin}x$z=cosx⇒dz=−sinx)

Using integration by parts,

$$\Rightarrow {e}^{y}\cdot {e}^{2\mathrm{cos}x}=\frac{1}{2}\cdot {z}^{2}\cdot {e}^{2z}-\frac{1}{2}z\cdot {e}^{2z}+\frac{{e}^{2z}}{4}+C$$⇒ey⋅e2cosx=21​⋅z2⋅e2z−21​z⋅e2z+4e2z​+C

$$\Rightarrow {e}^{y}\cdot {e}^{2\mathrm{cos}x}=\frac{1}{2}\cdot {\mathrm{cos}}^{2}x\cdot {e}^{2\mathrm{cos}x}-\frac{1}{2}\mathrm{cos}x\cdot {e}^{2\mathrm{cos}x}+\frac{{e}^{2\mathrm{cos}x}}{4}+C$$⇒ey⋅e2cosx=21​⋅cos2x⋅e2cosx−21​cosx⋅e2cosx+4e2cosx​+C

at $x=\frac{\pi }{2},y=0\Rightarrow C=\frac{3}{4}$x=2π​,y=0⇒C=43​

$$\Rightarrow {e}^{y}=\frac{1}{2}{\mathrm{cos}}^{2}x-\frac{1}{2}\mathrm{cos}x+\frac{1}{4}+\frac{3}{4}\cdot {e}^{-2\mathrm{cos}x}$$⇒ey=21​cos2x−21​cosx+41​+43​⋅e−2cosx

Put $x=0$x=0

$$\Rightarrow y=\mathrm{log}[\frac{1}{4}+\frac{3}{4}{e}^{-2}]\Rightarrow \alpha =\frac{1}{4},\beta =\frac{3}{4}$$⇒y=log[41​+43​e−2]⇒α=41​,β=43​

So, $\alpha +\beta =\frac{1}{4}+\frac{3}{4}=1$α+β=41​+43​=1