Question

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

$xdx+(y-2x)dy=0$

Short answer

The solution of the given differential equation is$\text{(y - x)ln(y - x) + x = C(y - x)}$.

Step-by-step solution

Step 1:Define substitution method for differentiation.

Often, the firststep 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 $\text{y = ux}$or $\text{x = vy}$, where $u$and localid="1654938340711" $v$ 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 for into the given differential equation.

Simplify further asshown below.

$$\begin{gathered} {\text{x[1 + u(u - 2)]dx + (u - 2)x}}^{\text{2}}\text{du = 0} \\ \text{x}\left[{\text{u}}^{\text{2}}\text{- 2u + 1}\right]{\text{dx + (u - 2)x}}^{\text{2}}\text{du = 0} \\ {\text{x(u - 1)}}^{\text{2}}{\text{dx + (u - 2)x}}^{\text{2}}\text{du = 0} \end{gathered}$$

Separate the variables and integrate the function.

Separate the variables and simplify it.

$\frac{\text{u - 2}}{{\text{(u - 1)}}^{\text{2}}}\text{du = -}\frac{\text{1}}{\text{x}}\text{dx}$

Integrate on both sides.

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

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

$$\begin{gathered} \text{ln}\left(\frac{\text{y}}{\text{x}}\text{- 1}\right)\text{+}\frac{\text{1}}{\frac{\text{y}}{\text{x}}\text{- 1}}\text{= - lnx + C} \\ \text{ln}\left[\text{x}\left(\frac{\text{y}}{\text{x}}\text{- 1}\right)\right]\text{+}\frac{\text{x}}{\text{y - x}}\text{= C} \\ \text{ln(y - x) +}\frac{\text{x}}{\text{y - x}}\text{= C} \\ \text{(y - x)ln(y - x) + x = C(y - x)} \end{gathered}$$

Thus, the solution of the given differential equation is $\text{(y - x)ln(y - x) + x = C(y - x)}$