Question

The solution of ${\sec }^{2}y\frac{dy}{dx}+2x\tan y=x^3$ is

• A. $\tan y=\frac{x^2}{2}+\frac{1}{2}+c{e}^{-x^2}$
• B. $\tan y=\frac{x^2}{2}-\frac{1}{2}+c{e}^{-x^2}$
• C. $\tan y$ $=\frac{x^2}{2}-\frac{1}{2}+c(e^x{)}^2$
• D. $\tan y=\frac{x^2}{2}+\frac{1}{2}$+$c(e^x{)}^2$

Answer 1

The correct option is C $\tan y$ $=\frac{x^2}{2}-\frac{1}{2}+c(e^x{)}^2$

$\begin{array}{c}{\sec }^{2}y\frac{dy}{dx}+2x\tan y=x^3\to \left(i\right)\\ let\tan y=t\to \left(ii\right)\\ differentequation\left(ii\right)w.r.tx\\ weget\\ {\sec }^{2}y\frac{dy}{dx}=\frac{dt}{dx}\\ nowequation\left(i\right)becomes\\ \frac{dt}{dx}+2xt+x^3\end{array}$

This is linear equation

$\begin{array}{c}I.F={\int }_{e}pdx={\int }_{e}2xdx=e^{x^2}\\ tx{e}^{x^2}=\int x^3\cdot e^{x^2}dx+c\\ t{e}^{x^2}=\int x\cdot x^2\cdot e^{x^2}dx+c\left[\begin{array}{c}put{x}^2=u\\ differenceweget\\ 2xdx=du\end{array}\right]\\ t{e}^{x^2}=\frac{1}{2}\int x\cdot e^{x^2}dx+c\left[bypartsinteger\right]\\ t{e}^{x^2}=\frac{1}{2}\left[x\cdot e^x-\int 1\cdot e^{x}dx\right]\\ t{e}^{x^2}=\frac{1}{2}\left[4e^x-e^x\right]+c\\ t{e}^{x^2}=\frac{e^{x^2}}{2}\left(x^2-1\right)+c\\ \tan y=\frac{1}{2}\left[x^2-1\right]+c\cdot e^{x^2}\end{array}$