Question

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

ydx = 2(x + y)dy

Short answer

The solutions of the given differential equation are $\left|x\right|=In\left|2\left(\frac{y}{x}\right)+1\right|-2In\left|\frac{y}{x}\right|+C$and $y=C_2\left(\frac{x}{y}+2\right).$

Step-by-step solution

Define substitution method for differentiation.

Often, the initial step in solving a differential equation is to use a substitution to change it into another differential equation.

The substitutions 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 = ux, where u and v are new dependent variables.

Substitution of variables by equations using property.

The given differential equation is a homogeneous function. So, let y = ux. Then,$\frac{dy}{dx}=u+x\frac{du}{dx}$.

Substitute them into the differential equation.

$\begin{array}{rcl}ydx-2\left(x-y\right)dy& =& 0\\ uxdx-2\left(x+ux\right)udx+xdu& =& 0\\ uxdx-2uxdx-2x^{2}du-2u^{2}xdx-2u{x}^{2}du& =& 0\\ -uxdx-2u^{2}xdx-2x^{2}du-2u{x}^{2}du& =& 0\\ \left(-u-2u^2\right)xdx-\left(2x^2\right)\left(du+udu\right)& =& 0\end{array}$

Integrate the function.

Let separate the variables be,

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

Integrate on both sides.

role="math" localid="1663840957679" $\begin{array}{rcl}\int \left(\frac{1}{2}{x}^{-1}dx\right)& =& \int \left(\frac{u+1}{-u-2u^2}\right)du\\ \frac{1}{2}In\left|x\right|& =& -\int \left(\frac{u+1}{2u^2+u}\right)du=-\int \left(\frac{A}{u}+\frac{B}{2u+1}\right)du\end{array}$

Using partial fractions, the integral becomes:

$\begin{array}{rcl}A\left(2u+1\right)+B\left(u\right)& =& u+1\\ 2Au+Bu+A& =& u+1\\ 2A+B& =& 1\\ A& =& 1,B=-1\end{array}$

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

Substitute the equation $v=2u+1$and $\frac{1}{2}dv=du$into the above equation.

$$\begin{gathered} \frac{1}{2}In\left|x\right|=\frac{1}{2}In\left|2u+1\right|-In\left|u\right|+C \\ y=ux \\ u=\frac{y}{x} \\ \frac{1}{2}In\left|x\right|=\frac{1}{2}In\left|2{\left(\frac{y}{x}\right)}^x+1\right|-In\left|\frac{y}{x}\right|+C \end{gathered}$$

Hence, the solution is, $\frac{1}{2}In\left|x\right|=\frac{1}{2}In\left|2{\left(\frac{y}{x}\right)}^x+1\right|-In\left|\frac{y}{x}\right|+C$.

Substitution of variables by equations using property.

The given differential equation is a homogeneous function. So, let x = vy. Then, $\frac{dx}{dy}=v+y\frac{dv}{dy}$.

Substitute them into the differential equation.

$\begin{array}{rcl}ydx-2\left(x+y\right)dy& =& 0\\ y\left(vdy+ydv\right)-2\left(vy+y\right)dy& =& 0\\ vydy+y^{2}dv-2vydy-2ydy& =& 0\end{array}$

Integrate the function.

Let separate the variables be,

$\begin{array}{rcl}{y}^{2}dv& =& \left(v+2\right)ydy\\ \frac{1}{v+2}dv& =& \frac{1}{y}dy\end{array}$

Integrate on both sides.

$$\begin{gathered} \int \frac{1}{v+2}dv=\int \frac{1}{y}dy \\ In\left|y\right|=In\left|v+2\right|+C \\ {e}^{In\left|y\right|}=e^{In\left|v+2\right|+C} \\ y=\left(v+2\right)\times e^c \end{gathered}$$

Substitute the equation$v=\frac{x}{y}$and $e^c=C_2$into the above equation.

Hence, the solution is, $y=C_2\left(\frac{x}{y}+2\right)$.