Question

Each DE in Problems $1-14$is homogeneous. In Problems$1-10$solve the given differential equation by using an appropriate substitution.$\left(y^2+yx\right)dx-x^{2}dx=0$

Short answer

The solution of the given differential equation is$y=\frac{x}{C-\ln \left|x\right|}$

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 u and$v$ are new dependent variables.

Substitution of variables by equations using property.

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

$dy=udx+xdu$.Substitute for and for into the given differential equation.

role="math" localid="1654844038531" $\begin{array}{rcl}\left({\left(ux\right)}^2+uxx\right)dx-x^2\left(udx+xdu\right)& =& 0\\ \left(u^2{x}^2\right)dx-x^2\left(udx+xdu\right)& =& 0\\ u^2{x}^{2}dx+u{x}^{2}dx-u{x}^{2}dx-x^{3}du& =& 0\\ u^2{x}^{2}dx& =& x^{3}du\\ & & \\ & & \\ & & \end{array}$

Simplify further as shown below.

$$\begin{gathered} \frac{x^2}{x^3}dx=\frac{1}{u^2}du \\ \frac{1}{x}du=\frac{1}{u^2}du \end{gathered}$$

 Step 3: Integrate the function.

Integrate on both sides.

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

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

$$\begin{gathered} \ln \left|x\right|+C=-\frac{1}{\left({\displaystyle \frac{Y}{X}}\right)} \\ \ln \left|x\right|+C=-\frac{x}{y} \\ \frac{y}{x}=\frac{1}{C-\ln \left|x\right|} \\ y=\frac{x}{C-\ln \left|x\right|} \end{gathered}$$

Thus, the solution of the given differential equation is $y=\frac{x}{C-\ln \left|x\right|}$