Question

Each DE in Problems $1-14$is homogeneous. In Problems $1-10$solve the given differential equation by using an appropriate substitution.

$\frac{dy}{dx}=\frac{x+3}{3x+y}$

Short answer

The solution of the given differential equation is ${\left(x-y\right)}^2=C\left(x-y\right)$

Step-by-step solution

Step 1:Define substitution method for differentiation.

Often, the first stepin 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 $dy$into the given differential equation.

$$\begin{gathered} \frac{dy}{dx}=\frac{x+3y}{3x+y} \\ \left[u+x\frac{du}{dx}\right]=\frac{x+3\left[ux\right]}{3x+\left[ux\right]} \\ u+x\frac{du}{dx}=\frac{x+3ux}{3x+ux} \end{gathered}$$

Simplify further as shown below.

$$\begin{gathered} x\frac{du}{dx}=\frac{x\left(1+3u\right)}{x\left(3+u\right)}-u \\ x\frac{du}{dx}=\frac{1+3u}{3+u}-\frac{u\left(3+u\right)}{3+u} \\ x\frac{du}{dx}=\frac{1+3u-3u-u^2}{3+u} \\ x\frac{du}{dx}=\frac{1-u^2}{3+u} \end{gathered}$$

 Step 3: Integrate the function.

Separate the variables and simplify it.

$\frac{3+u}{1-u^2}du=\frac{1}{x}dx$

Use the partial decomposition for the above equation.

$\left(\frac{2}{1-u}+\frac{1}{1+u}\right)du=\frac{1}{x}du$

Integrate on both sides.

localid="1654923711940" $\begin{array}{rcl}\int \left(\frac{2}{1-u}+\frac{1}{1+u}\right)du& =& \int \frac{1}{x}dx\\ -2\ln \left(1-u\right)+\ln \left(1+u\right)& =& \ln x+\ln C\\ \ln \left[\frac{1+u}{{\left(1-u\right)}^2}\right]& =& \ln Cx\end{array}$

Use exponential on both sides.

$\frac{1+u}{(1-u{\}}^2}=Cx$

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

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

Thus, the solution of the given differential equation is ${\left(x-y\right)}^2=C\left(x+y\right)$