Question

Solve the given differential equation by substitution method:

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\left(x+y\right)$

Short answer

We will solve the given differential equation by substituting.

Step-by-step solution

definition of solving differential equation by substitution

Often the first step in solving the differential equation consists of transforming it into another differential equation by means of a substantial.

For example: Suppose we wish to transform the first order differential equation dy/dx=f(x,y) by the substitution y=g(x,u), where u is regarded as a function of the variable x.

Step 2:

Let u=x+y

$$\begin{gathered} u=x+y \\ \frac{du}{dx}=1+\frac{dy}{dx} \\ \frac{dy}{dx}=\frac{du}{dx}-1 \end{gathered}$$

Do the substitution:

$$\begin{gathered} \left[\frac{du}{dx}-1\right]=\sin u \\ \frac{du}{dx}=\sin u+1 \end{gathered}$$

Step 3:

Solve by separation of variables :

$$\begin{gathered} \frac{1}{1+\sin u}du=dx \\ \int \frac{1}{1+\sin u}du=\int dx \\ \int \frac{1}{1+\sin u}\frac{1-\sin u}{1-\sin u}du=\int dx \\ \int \frac{1-\sin u}{1-{\sin }^{2}u}du=\int dx \\ \int \frac{1-\sin u}{{\cos }^{2}u}du=\int dx \\ \int \left(\frac{1}{{\cos }^{2}u}-\frac{\sin u}{{\cos }^{2}u}\right)du=\int dx \\ \int \left(se{c}^{2}u-\tan usecu\right)du=\int dx \\ \tan u-secu=x+C \end{gathered}$$

Step 4:

Substitute back u=x+y

$\tan \left(x+y\right)-sec\left(x+y\right)=x+C$

Hence, the final result is;

$\tan \left(x+y\right)-sec\left(x+y\right)=x+C$