Differential Equations — Maths STD 12 Science — Question

$\text{dy}=\text{x}^5\tan^{-1}(\text{x}^3)\text{dx}$

$\int\text{dy}=\int\text{x}^5\tan^{-1}(\text{x}^3)\text{dx}$

put $\text{x}^3=\text{t}$

$\Rightarrow3\text{x}^2\text{dx}=\text{dt}$

$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$

So,

$\int\text{dy}=\frac{1}{3}\Big[\tan^{-1}\text{t}\int\text{t dt}=\int\Big(\frac{1}{1+\text{t}^2}\Big)\times\int\text{t dx}\Big)\Big]\text{dt}+\text{C}$

Using integration by parts

$\text{y}=\frac{1}{3}\Big[\frac{\text{t}^2}{2}+\tan^{-1}-\int\frac{\text{t}^2}{2(\text{t}^2+1)}\text{dt}\Big]+\text{C}$

$=\frac{1}{6}\text{t}^2\tan^{-1}\text{t}-\frac{1}{6}\int\Big(\frac{\text{t}^2}{\text{t}^2+1}\Big)\text{dt}+\text{C}$

$\text{y}=\frac{1}{6}\text{t}^2\tan^{-1}\text{t}-\frac{1}{6}\int\Big(1-\frac{1}{\text{t}^2+1}\Big)\text{dt}+\text{C}$

$=\frac{1}{6}\text{t}^2\tan^{-1}\text{t}-\frac{1}{6}\text{t}+\frac{1}{6}\tan^{-1}\text{t}+\text{C}$

$\text{y}=\frac{1}{6}(\text{t}^2+1)\tan^{-1}\text{t}-\frac{1}{6}\text{t}+\text{C}$

$\text{y}=\frac{1}{6}[(\text{t}^2+1)\tan^{-1}\text{t}-\text{t}]+\text{C}$

So,

$\text{y}=\frac{1}{6}[(\text{x}^6+1)\tan^{-1}(\text{x}^3)-\text{x}^3]+\text{C}$