Question

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

$-ydx+\left(x+\sqrt{xy}\right)dy=0$

Short answer

The solution of the given differential equation is$4x=y{\left(\ln y+C\right)}^2$.

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,$x=vy$ if,then.

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

$$\begin{gathered} -ydx+\left(x+\sqrt{xy}\right)dy=0 \\ -y\left[vdy+ydv\right]+\left(\left[vy\right]+\sqrt{\left[vy\right]y}\right)dy=0 \\ -vydy-y^{2}dv+vydy+y\sqrt{v}dy=0 \\ -y^{2}dv+y\sqrt{v}dy=0 \\ -ydv+\sqrt{v}dy=0 \end{gathered}$$

Separate the variables and integrate the function.

Separate the variables and simplify it.

$\frac{1}{\sqrt{v}}dv=\frac{1}{y}dy$

Integrate on both sides.

$\begin{array}{rcl}\int \frac{1}{\sqrt{v}}dv& =& \int \frac{1}{y}dy\\ 2\sqrt{v}& =& \ln y+C\end{array}$

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

$\begin{array}{rcl}2\sqrt{\frac{x}{y}}& =& \ln y+C\\ 4\frac{x}{y}& =& {\left(\ln y+C\right)}^2\\ 4x& =& y{\left(\ln y+C\right)}^2\end{array}$

Thus, the solution of the given differential equation is $4x=y{\left(\ln y+C\right)}^2$