Question

In Problems 1–22, solve the given differential equation by separation of variables.

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{xy + 2y - x - 2}}{\text{xy - 3y + x - 3}}$

Short answer

$e^y(y-1{)}^2=C{e}^x(x-3{)}^5$

Step-by-step solution

Definition of "separable equation"

A first-order differential equation of the form$\frac{dy}{dx}=g\left(x\right)h\left(y\right)$ is said to be separable or to have separable variables.

Grouping factor and separating the variables

$\frac{dy}{dx}=\frac{xy+2y-x-2}{xy-3y+x-3}$

Factor the numerator and denominator into the right hand side.

$$\begin{gathered} \frac{dy}{dx}=\frac{y(x+2)-(x+2)}{y(x-3)+(x-3)} \\ \frac{dy}{dx}=\frac{(x+2)(y-1)}{(x-3)(y+1)} \end{gathered}$$

Separate the variables.

$$\begin{gathered} \left(x-3\right)\left(y+1\right)dy=\left(x+2\right)\left(y-1\right)dx \\ \int \frac{y+1}{y-1}dy=\int \frac{x+2}{x-3}dx \\ \int \frac{y-1+2}{y-1}dy=\int \frac{x-3+5}{x-3}dx \end{gathered}$$

Integration and simplification

Distribute the numerators and simplify

$$\begin{gathered} \int \left(\frac{y-1}{y-1}+\frac{2}{y-1}\right)dy=\int \left(\frac{x-3}{x-3}+\frac{5}{x-3}\right)dx \\ \int \left(1+\frac{2}{y-1}\right)dy=\int \left(1+\frac{5}{x-3}\right)dx \\ \int dy+2\int \frac{1}{y-1}dy=\int dx+5\int \frac{1}{x-3}dx \\ y+2\ln |y-1|=x+5\ln |x-3|+c \\ y+\ln |y-1{|}^2=x+\ln |x-3{|}^5+c \end{gathered}$$

Take to both sides

$e^{y+\ln |y-1{|}^2}=e^{x+\ln |x-3{|}^5+c}$

Remember that$e^{a+b}=e^a{e}^b$

$e^y{e}^{\ln |y-1{|}^2}=e^x{e}^c{e}^{\ln |x-3{|}^5}$

So,

$e^y(y-1{)}^2=C{e}^x(x-3{)}^5$

Hence, the solution is${\text{e}}^{\text{y}}{\text{(y - 1)}}^{\text{2}}{\text{= Ce}}^{\text{x}}{\text{(x - 3)}}^{\text{5}}$