Question

In Problems 19-26 solve the given differential equation.

$\mathbf{(}{\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}\mathit{d}\mathit{x}\mathbf{=}\mathit{y}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathit{x}\mathit{d}\mathit{y}$

Short answer

The solution of the given differential equation is $2x+\sin 2x=2\ln |y^2+1|+C$.

Step-by-step solution

Note the given data

Given the initial value problem .$(y^2+1)dx=y{\sec }^{2}xdy$

We know that, $2{\cos }^{2}x=1+\cos 2x$.

Calculation part of the differential equation

Solving the differential equation by separated method as follows:

$\begin{array}{c}\frac{dx}{{\sec }^{2}x}=\frac{y}{y^2+1}dy\\ \int \frac{dx}{{\sec }^{2}x}=\int \frac{y}{y^2+1}dy\\ \int {\cos }^{2}xdx=\int \frac{y}{y^2+1}dy\end{array}$

Putting ${\cos }^{2}x=\frac{1}{2}\left(1+\cos 2x\right)$and simplify as:

$\begin{array}{c}\int \frac{1}{2}\left(1+\cos 2x\right)dx=\int \frac{y}{y^2+1}dy\\ \frac{1}{2}\int \left(1+\cos 2x\right)dx=\int \frac{y}{y^2+1}dy\\ \int \left(1+\cos 2x\right)dx=2\int \frac{y}{y^2+1}dy\\ \int \left(1+\cos 2x\right)dx=\int \frac{d(y^2+1)}{y^2+1}\end{array}$

Further simplify the equation as follows:

$\begin{array}{c}x+\frac{\sin 2x}{2}=\ln |y^2+1|+c_1\\ 2x+\sin 2x=2\ln |y^2+1|+2c_1\\ 2x+\sin 2x=2\ln |y^2+1|+C\end{array}$

where $C=2c_1$ is the constant of integration.

Hence, the solution of the given differential equation is $2x+\sin 2x=2\ln |y^2+1|+C$.