Let the solution curve y = y ( x ) of the differential equation d y d x - 3 x 5 tan - 1 x 3 1 + x 6 3 2 y =…

Given differential equation is$\frac{dy}{dx}+\left(\frac{-3x^5{\tan }^{-1}\left(x^3\right)}{{\left(1+x^6\right)}^{3/2}}\right)y=2x{e}^{\left\{\frac{x^3-{\tan }^{-1}{x}^3}{\sqrt{{\left(1+x\right)}^6}}\right\}}$

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

Hence,

$\mathrm{I}.\mathrm{F}.=e^{\int \frac{-3x^5{\tan }^{-1}{x}^3}{{\left(1+x^6\right)}^{3/2}}dx}$

Put ${\tan }^{-1}{x}^3=u\Rightarrow \left(\frac{3x^2}{1+x^6}\right)dx=du$

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

$\Rightarrow \mathrm{I}.\mathrm{F}.=e^{-\int \frac{u\tan u}{\sec u}du}$

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

$\Rightarrow \mathrm{I}.\mathrm{F}.=e^{-\left[-u\cos u+\int \cos udu\right]}$

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

$\Rightarrow \mathrm{I}.\mathrm{F}.=e^{\left\{{\tan }^{-1}\left(x^3\right)\left(\frac{1}{\sqrt{1+x^6}}\right)-\left(\frac{x^3}{\sqrt{1+x^6}}\right)\right\}}$

$\Rightarrow \mathrm{I}.\mathrm{F}.=e^{\left\{\frac{{\tan }^{-1}{x}^3-x^3}{\sqrt{1+x^6}}\right\}}$

Solution of differential equation is

$y·e^{\frac{{\tan }^{-1}{x}^3-x^3}{\sqrt{1+x^6}}}=\int 2x{e}^{\left(\frac{x^3-{\tan }^{-1}{x}^3}{\sqrt{1+x^6}}\right)}·e^{\left(\frac{{\tan }^{-1}\left(x^3\right)-x^3}{\sqrt{1+x^6}}\right)}dx$

$\Rightarrow y·e^{\left\{\frac{{\tan }^{-1}{x}^3-x^3}{\sqrt{1+x^6}}\right\}}=\int 2xdx$

$\Rightarrow y·e^{\left\{\frac{{\tan }^{-1}{x}^3-x^3}{\sqrt{1+x^6}}\right\}}=x^2+C$

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

$y{e}^{\left\{\frac{{\tan }^{-1}{x}^3-x^3}{\sqrt{1+x^6}}\right\}}=x^2$

Now, put $x=1$, then we get

$y\left(1\right)e^{\frac{{\tan }^{-1}\left(1\right)-1}{\sqrt{2}}}=1$

$\Rightarrow y\left(1\right)e^{\frac{\mathrm{\pi }-4}{4\sqrt{2}}}=1$

$\Rightarrow y\left(1\right)=\frac{1}{e^{\left(\frac{\mathrm{\pi }-4}{4\sqrt{2}}\right)}}$

$\Rightarrow y\left(1\right)=e^{\frac{4-\pi }{4\sqrt{2}}}$