Question

Each DE in Problems $1-14$is homogeneous. In Problems $1-10$ solve the given differential equation by using an appropriate substitution$\frac{dy}{dx}=\frac{y-x}{y+x}$.

Short answer

The solution of the given differential equation is$\ln \left(x^2+y^2\right)+2{\tan }^{-1}\left(\frac{y}{x}\right)=C$.

Step-by-step solution

Step 1:Define substitution method for differentiation.

Often, the first step in solving a differential equation is to use a substitution to change it into another differential equation. Thesubstitutions that can be employed to solve a homogeneous differential equation are suggested by their properties. A homogeneous equation can be reduced to a separable first-order differential equation by substituting $y=ux$or $x=vy$, where and are new dependent variables.

Substitution of variables by equations using property.

The given differential equation is a homogeneous function. So, $y=ux$if,then

$\frac{dy}{dx}=u+x\frac{du}{dx}$

Substitute $ux$for $y$and $udx+xdu$for into the given differential equation.

$$\begin{gathered} \frac{dy}{dx}=\frac{y-x}{y+x} \\ \left(u+x\frac{du}{dx}\right)=\frac{ux-x}{ux+x} \\ x\frac{du}{dx}=\frac{ux-x}{ux+x}-u \end{gathered}$$

Simplify further as shown below.

$$\begin{gathered} x\frac{du}{dx}=\frac{ux-x}{ux+x}-\frac{u\left(ux+x\right)}{ux+x} \\ x\frac{du}{dx}=\frac{ux-x-u^{2}x-ux}{ux+x} \\ x\frac{du}{dx}=\frac{-u^{2}x-x}{ux+x} \end{gathered}$$

Simplify the above equation further as shown below.

$$\begin{gathered} x\frac{du}{dx}=\frac{-x\left(u^2+1\right)}{x\left(u+1\right)} \\ x\frac{du}{dx}=\frac{-\left(u^2+1\right)}{\left(u+1\right)} \end{gathered}$$

Integrate the function

Separate the variables and simplify it.

$\frac{u+1}{u^2+1}du=\frac{1}{x}dx$

Integrate on both sides.

$$\begin{gathered} \int \left(\frac{u}{u^2+1}+\frac{1}{u^2+1}\right)du=-\int \frac{1}{x}dx \\ \frac{1}{2}\ln \left(u^2+1\right)+{\tan }^{-1}u=\ln x+C \end{gathered}$$

perform substitution

Substitute $\frac{y}{x}$for into the above equation.

$$\begin{gathered} \frac{1}{2}\ln \left({\left(\frac{y}{x}\right)}^2+1\right)+{\tan }^{-1}\left(\frac{y}{x}\right)=\ln x+C \\ \ln \left({\left(\frac{y}{x}\right)}^2+1\right)+2{\tan }^{-1}\left(\frac{y}{x}\right)+2\ln x=C \end{gathered}$$

Simplify further asshown below.

localid="1654861089605" $$\begin{gathered} \ln \left({\left(\frac{y}{x}\right)}^2+1\right)+2{\tan }^{-1}\left(\frac{y}{x}\right)+2\ln x=C \\ \ln \left(x^{2+y^2}\right)+2{\tan }^{-1}\left(\frac{y}{x}\right)=c \end{gathered}$$

Thus, the solution of the given differential equation is

$\ln \left(x^2+y^2\right)+2{\tan }^{-1}\left(\frac{y}{x}\right)=C$