Question

Find a solution of $\mathbf{x}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}{\mathbf{y}}^2\mathbf{-}\mathbf{y}$that passes through the indicated points.

(a)$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}$

(b)$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$

(c)role="math" localid="1667815116761" $(\frac{1}{2},\frac{1}{2})$

(d)role="math" localid="1667815153576" $(2,\frac{1}{4})$

Short answer

1. The equation is $y=1$.
2. The equation is $y=0$.
3. The equation is$y=\frac{1}{1+2x}.$
4. The equation is $y=\frac{2}{2+3x}$.

Step-by-step solution

Definition

A differential equation is an equation with one or more derivatives of a function.

Simplify differential equation

First find the general solution of the differential equation using separable equation.

$\begin{array}{c}x\frac{dy}{dx}=y^2-y\\ \frac{dy}{y^2-y}=\frac{dx}{x}\\ \frac{dy}{y(y-1)}=\frac{dx}{x}\end{array}$

Use partial fraction

Use the partial fractions to solve $\frac{1}{y(y-1)}$as,

$\begin{array}{c}\frac{1}{\mathrm{y}(\mathrm{y}-1)}=\frac{\mathrm{A}}{\mathrm{y}}+\frac{\mathrm{B}}{\mathrm{y}-1}\\ \frac{1}{\mathrm{y}(\mathrm{y}-1)}=\frac{\mathrm{A}(\mathrm{y}-1)+\mathrm{By}}{\mathrm{y}(\mathrm{y}-1)}\\ 1=(\mathrm{A}+\mathrm{B})\mathrm{y}-\mathrm{A}\end{array}$

Comparing both sides we have:

$\begin{array}{l}\Rightarrow (\mathrm{A}+\mathrm{B})=0,-\mathrm{A}=1\\ \Rightarrow \mathrm{A}=-1,\mathrm{B}=-\mathrm{A}=1\end{array}$

Integrate

Therefore, the integral,$\frac{1}{y(y-1)}$ becomes$\frac{1}{\mathrm{y}(\mathrm{y}-1)}=-\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{y}-1}$

Hence, the integration is given by,

$\left[\frac{-1}{\mathrm{y}}+\frac{1}{\mathrm{y}-1}\right]\mathrm{dy}=\frac{\mathrm{dx}}{\mathrm{x}}$

Integrate on both sides to obtain the following:

Simplifying integral

$\begin{array}{c}\ln \left|\frac{y-1}{y}\right|=\ln \left|cx\right|\\ \frac{y-1}{y}=cx\\ y-1=ycx\\ y-ycx=1\\ y(1-cx)=1\\ y=\frac{1}{1-cx}\end{array}$

Thus, $y\left(x\right)=\frac{1}{1-cx}$is the general solution of given differential equation.

Now, find $\frac{dy}{dx}=0$.

$\begin{array}{c}{y}^2-y=0\\ y(y-1)=0\\ y=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y=1\end{array}$

Check for options

For option (a)

$y=1$is a constant (equilibrium) solution corresponding to $y\left(0\right)=1$.

For option (b)

$y=0$is a constant (equilibrium) solution corresponding to$y\left(0\right)=0$ doesn't correspond to a value of $c$).

For option(c)

The equation passes through the point $\left(\frac{1}{2},\frac{1}{2}\right)$.

Substitute $x=\frac{1}{2}$and $y=\frac{1}{2}$in $y=\frac{1}{1-cx}$to obtain the $c$value.

$\begin{array}{l}\frac{1}{2}=\frac{1}{1-\mathrm{c}\left(\frac{1}{2}\right)}\\ \frac{1}{2}=\frac{1}{\frac{2-\mathrm{c}}{2}}\\ \frac{1}{2}=\frac{2}{2-\mathrm{c}}\\ 2-\mathrm{c}=4\\ \mathrm{c}=-2\\ \text{Thus,thesolutionof}\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{y}}^2-\mathrm{y}\text{passingthrough}\left(\frac{1}{2},\frac{1}{2}\right)\text{is}\mathrm{y}=\frac{1}{1+2\mathrm{x}}.\end{array}$

Check for option

The equation passes through the point$\left(2,\frac{1}{4}\right)$.

Substitute $x=2$and $y=\frac{1}{4}$in$y=\frac{1}{1-cx}$ to obtain the $c$value.

$\begin{array}{c}\frac{1}{4}=\frac{1}{1-c\left(2\right)}\\ \frac{1}{4}=\frac{1}{1-2c}\\ 1-2c=4\\ -2c=3\\ c=-\frac{3}{2}\end{array}$

Therefore,

$\begin{array}{c}y=\frac{1}{1+\frac{3}{2}x}\\ =\frac{1}{\frac{2+3x}{2}}\\ =\frac{2}{2+3x}\end{array}$

Thus, the solution of $x\frac{dy}{dx}=y^2-y$passing through$\left(2,\frac{1}{4}\right)$ is $y=\frac{2}{2+3x}$.