Question

Solve the differential equation $\frac{dy}{dx}=x^{5}ta{n}^{-1}\left(x^3\right)$.

Answer 1

$\begin{array}{c}\frac{dy}{dx}=x^5{\tan }^{-1}{x}^3\\ dy=\left(x^5{\tan }^{-1}{x}^3\right)dx\\ Takingintegrationonbothsides\\ \int dy=\int \left(x^5{\tan }^{-1}{x}^3\right)dx\\ Let{x}^3=t\\ 3x^{2}dx=dt\\ y=\frac{1}{3}\int t{\tan }^{-1}tdt+C\\ =\frac{1}{3}{\tan }^{-1}t\int tdt-\frac{1}{3}\int \left[\frac{d\left({\tan }^{-1}t\right)}{dx}\int tdt\right]dt+C\\ =\frac{t^2}{6}{\tan }^{-1}t-\frac{1}{6}\int \left[\frac{t^2+1-1}{1+t^2}\right]dt+C\\ =\frac{t^2}{2}{\tan }^{-1}t-\frac{1}{6}\int \left[1-\frac{1}{1+t^2}\right]dt+C\\ =\frac{t^2}{2}{\tan }^{-1}t-\frac{1}{6}\left(t-{\tan }^{-1}t\right)+C\\ y=\frac{x^6}{6}{\tan }^{-1}{x}^3-\frac{x^3}{6}+\frac{1}{6}{\tan }^{-1}\left(x^3\right)+C.\end{array}$