Question

(a) Use a CAS and the concept of level curves to plot representative graphs of members of the family of solutions of the differential equation $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{x}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{)}}{\mathbf{y}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{+}\mathbf{y}\mathbf{)}}$. Experiment with different numbers of level curves as well as various rectangular regions in the -plane until your result resembles Figure 2.2.6.

(b) On separate coordinate axes, plot the graph of the implicit solution corresponding to the initial condition$\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}$. Use a colored pencil to mark off that segment of the graph that corresponds to the solution curve of a solution $\varphi$that satisfies the initial condition. With the aid of a rootfinding application of a CAS, determine the approximate largest interval of definition of the solution$\varphi$. [Hint: First find the points on the curve in part (a) where the tangent is vertical.]

(c) Repeat part (b) for the initial condition$\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$.

Short answer

Answer: (a) So, the solution of the differential equation is$-y^2+\frac{1}{3}{y}^3=\frac{1}{2}{x}^2-\frac{1}{3}{x}^3+c$. (b) So, the solution curve of a solution is$-y^2+\frac{1}{3}{y}^3-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3=c$and the graph of the implicit solution is shown as below:

And the solution of part (c) is along with the graph

Step-by-step solution

Given Data

Consider the differential equation

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

Separate the variables by cross multiplying.

$y(-2+y)dy=x(1-x)dx$

Evaluation

Now we integrate the left hand side in terms of and the right hand side in terms of x.

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

Therefore, the solution is $-y^2+\frac{1}{3}{y}^3-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3=c$.

Draw graph

Below we plot the level curves corresponding to equation (2) for a variety of constant values$c=-105,-65,-35,-25,-15,-14,-1,-0,5,-0,1,2,5,10$. The graphing window is$-4⩽x⩽4$and$-3⩽y⩽5$.

(b)Step 1: Evaluation

Substitute the initial condition$y\left(0\right)=\frac{3}{2}$ into equation (1) in order to solve for.

$$\begin{gathered} c=-{\left(\frac{3}{2}\right)}^2+\frac{1}{3}{\left(\frac{3}{2}\right)}^3-\frac{1}{2}(0{)}^2+\frac{1}{3}(0{)}^3 \\ c=-\frac{9}{4}+\frac{9}{8} \\ c=-\frac{9}{8} \end{gathered}$$

Therefore, the implicit solution is $-y^2+\frac{1}{3}{y}^3-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3=-\frac{9}{8}$.

Substitution

In order to find the interval of definition for solution (3) we need to calculate the point(s) where the differential equation is undefined, that is the tangent is vertical. This happens when the denominator on the right hand side of equation (1) is zero.

$y(-2+y)=0$

So $y=0ory=2$. We substitute each solution of into equation (3) and solve for. For we have,

$$\begin{gathered} -\frac{9}{8}=-(0{)}^2+\frac{1}{3}(0{)}^3-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-\frac{9}{8} \\ =-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3 \end{gathered}$$

We use a CAS to get the estimation,$xz\approx 1.132$ . $y=2$For we have,

$$\begin{gathered} -\frac{9}{8}=-(2{)}^2+\frac{1}{3}(2{)}^3-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-\frac{9}{8} \\ =-4+\frac{8}{3}-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-\frac{9}{8} \\ =-4+\frac{8}{3}-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3\frac{59}{24} \\ =-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3 \end{gathered}$$

We use a CAS to get the estimation,$x\approx 1.713$.

Therefore, the largest interval of definition is$-1.132⩽x<1.713$.

Draw the graph

Below we plot the solution with the two red diamond points defining the interval of definition for the solution and the blue diamond point defining the initial condition.

(c)Step 1: Evaluation

Substitute the initial condition $y\left(0\right)=-2$into equation (1) in order to solve for .

$$\begin{gathered} c=-(-2{)}^2+\frac{1}{3}(-2{)}^3-\frac{1}{2}(0{)}^2+\frac{1}{3}(0{)}^3 \\ c=-4-\frac{8}{3} \\ c=-\frac{20}{3} \end{gathered}$$

Therefore, the implicit solution is$-y^2+\frac{1}{3}{y}^3-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3=-\frac{20}{3}$.

Interval of definition for solution

We need to calculate the point(s) where the differential equation is undefined, that is the tangent is vertical. This happens when the denominator on the right hand side of equation (1) is zero.

$y(-2+y)=0$

So $y=0ory=2$We substitute each solution of into equation (4) and solve for. For $y=0$we $$\begin{gathered} -\frac{20}{3}=-(0{)}^2+\frac{1}{3}(0{)}^3-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3 \\ -\frac{20}{3}=-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3 \end{gathered}$$have,

We use a CAS to get the estimation, $x\approx -2.296$. For $y=0$we have,

$$\begin{gathered} -\frac{20}{3}=-(2{)}^2+\frac{1}{3}(2{)}^3-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3 \\ -\frac{20}{3}=-4+\frac{8}{3}-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3 \\ -\frac{20}{3}=-4+\frac{8}{3}-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3 \\ -\frac{16}{3}=-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3 \end{gathered}$$

We use a CAS to get the estimation,$x\approx -2.106$.

Therefore, the largest interval of definition is $-2.106<x<\infty$.

Draw the graph

Below we plot the solution with the two red diamond points defining the location of vertical tangent lines for the solution and the blue diamond point defining the initial condition.