JEE Main 2023 — Mathematics Calculus

Given differential equation is$\frac{dy}{dx}+(\frac{-3{x}^{5}{\mathrm{tan}}^{-1}({x}^{3})}{{(1+{x}^{6})}^{3/2}})y=2x{e}^{{\frac{{x}^{3}-{\mathrm{tan}}^{-1}{x}^{3}}{\sqrt{{(1+x)}^{6}}}}}$dxdy​+((1+x6)3/2−3x5tan−1(x3)​)y=2xe(1+x)6​x3−tan−1x3​

This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$dxdy​+Py=Q.

Hence,

$$I.F.={e}^{\int \frac{-3{x}^{5}{\mathrm{tan}}^{-1}{x}^{3}}{{(1+{x}^{6})}^{3/2}}dx}$$I.F.=e∫(1+x6)3/2−3x5tan−1x3​dx

Put ${\mathrm{tan}}^{-1}{x}^{3}=u\Rightarrow (\frac{3{x}^{2}}{1+{x}^{6}})dx=du$tan−1x3=u⇒(1+x63x2​)dx=du

$$I.F.={e}^{-\int \frac{u\mathrm{tan}u}{\sqrt{1+{\mathrm{tan}}^{2}u}}du}$$I.F.=e−∫1+tan2u​utanu​du

$$\Rightarrow I.F.={e}^{-\int \frac{u\mathrm{tan}u}{\mathrm{sec}u}du}$$⇒I.F.=e−∫secuutanu​du

$$\Rightarrow I.F.={e}^{-\int u\mathrm{sin}udu}$$⇒I.F.=e−∫usinudu

$$\Rightarrow I.F.={e}^{-[-u\mathrm{cos}u+\int \mathrm{cos}udu]}$$⇒I.F.=e−[−ucosu+∫cosudu]

$$\Rightarrow I.F.={e}^{u\mathrm{cos}u-\mathrm{sin}u}$$⇒I.F.=eucosu−sinu

$$\Rightarrow I.F.={e}^{{{\mathrm{tan}}^{-1}({x}^{3})(\frac{1}{\sqrt{1+{x}^{6}}})-(\frac{{x}^{3}}{\sqrt{1+{x}^{6}}})}}$$⇒I.F.=etan−1(x3)(1+x6​1​)−(1+x6​x3​)

$$\Rightarrow I.F.={e}^{{\frac{{\mathrm{tan}}^{-1}{x}^{3}-{x}^{3}}{\sqrt{1+{x}^{6}}}}}$$⇒I.F.=e1+x6​tan−1x3−x3​

Solution of differential equation is

$$y\cdot {e}^{\frac{{\mathrm{tan}}^{-1}{x}^{3}-{x}^{3}}{\sqrt{1+{x}^{6}}}}=\int 2x{e}^{(\frac{{x}^{3}-{\mathrm{tan}}^{-1}{x}^{3}}{\sqrt{1+{x}^{6}}})}\cdot {e}^{(\frac{{\mathrm{tan}}^{-1}({x}^{3})-{x}^{3}}{\sqrt{1+{x}^{6}}})}dx$$y⋅e1+x6​tan−1x3−x3​=∫2xe(1+x6​x3−tan−1x3​)⋅e(1+x6​tan−1(x3)−x3​)dx

$$\Rightarrow y\cdot {e}^{{\frac{{\mathrm{tan}}^{-1}{x}^{3}-{x}^{3}}{\sqrt{1+{x}^{6}}}}}=\int 2xdx$$⇒y⋅e1+x6​tan−1x3−x3​=∫2xdx

$$\Rightarrow y\cdot {e}^{{\frac{{\mathrm{tan}}^{-1}{x}^{3}-{x}^{3}}{\sqrt{1+{x}^{6}}}}}={x}^{2}+C$$⇒y⋅e1+x6​tan−1x3−x3​=x2+C

Also, it passes through the origin then $C=0$C=0, hence

$$y{e}^{{\frac{{\mathrm{tan}}^{-1}{x}^{3}-{x}^{3}}{\sqrt{1+{x}^{6}}}}}={x}^{2}$$ye1+x6​tan−1x3−x3​=x2

Now, put $x=1$x=1, then we get

$$y(1){e}^{\frac{{\mathrm{tan}}^{-1}(1)-1}{\sqrt{2}}}=1$$y(1)e2​tan−1(1)−1​=1

$$\Rightarrow y(1){e}^{\frac{\pi -4}{4\sqrt{2}}}=1$$⇒y(1)e42​π−4​=1

$$\Rightarrow y(1)=\frac{1}{{e}^{(\frac{\pi -4}{4\sqrt{2}})}}$$⇒y(1)=e(42​π−4​)1​

$$\Rightarrow y(1)={e}^{\frac{4-\pi }{4\sqrt{2}}}$$⇒y(1)=e42​4−π​