Question

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

$(x+y)dx+xdy=0$

Short answer

The solution of the given differential equation is$\text{y =}\frac{\text{C}}{\text{2x}}\text{-}\frac{\text{x}}{\text{2}}$

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$\text{y = ux}$or $\text{x = vy}$, where $u$and $v$ are new dependent variables.

Substitution of variables by equations using property.

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

$\text{dy = udx + xdu}$

Substitute $\text{ux}$for $\text{y}$, and $\text{udx + xdu}$for into the given differential equation.

$$\begin{gathered} \text{(x +}\underset{\text{y}}{\underbrace{\text{ux}}}\text{)dx + x}\underset{\text{dy}}{\underbrace{\text{(udx + xdu)}}}\text{= 0} \\ {\text{xdx + uxdx + uxdx +x}}^{\text{2}}\text{du = 0} \\ {\text{xdx + uxdx + uxdx +x}}^{\text{2}}\text{du = 0} \\ {\text{xdx + 2uxdx +x}}^{\text{2}}\text{du = 0} \end{gathered}$$

Simplify further as shown below.

${\text{(1 + 2u)xdx +x}}^{\text{2}}\text{du = 0}$

Separate the variables and integrate the function.

Separate the variables and simplify it.

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

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

Integrate on both sides.

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

Simplify further

Take exponential on both sides of the resulting equation

$$\begin{gathered} {\text{e}}^{\text{ln|2u + 1|}}{\text{=e}}^{\text{- 2ln|x| + C}} \\ {\text{e}}^{\text{ln|2u + 1|}}{\text{=e}}^{\text{c}}{\text{e}}^{\text{- 2ln|x|}} \\ {\text{e}}^{\text{ln|2u + 1|}}{\text{= Ce}}^{\text{ln}\left({\text{x}}^{\text{- 2}}\right)} \\ {\text{2u + 1 = Cx}}^{\text{- 2}} \end{gathered}$$

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

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

Solve for $y$.

$$\begin{gathered} \text{2y =}\frac{\text{C}}{\text{x}}\text{- x} \\ \text{y =}\frac{\text{C}}{\text{2x}}\text{-}\frac{\text{x}}{\text{2}} \end{gathered}$$

Thus, the solution of the given differential equation is $\text{y =}\frac{\text{C}}{\text{2x}}\text{-}\frac{\text{x}}{\text{2}}$