Question

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 $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{-}\frac{\mathbf{8}\mathbf{x}\mathbf{+}\mathbf{5}}{\mathbf{3}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}$. Experiment with different numbers of level curves as well as various rectangular regions defined by$\mathbf{a}⩽\mathbf{x}⩽\mathbf{b}\mathbf{,}\mathbf{c}⩽\mathbf{y}⩽\mathbf{d}\mathbf{.}$

(b) On separate coordinate axes plot the graphs of the particular solutions corresponding to the initial conditions:$\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{2}\mathbf{;}\mathbf{y}\left(\mathbf{-}\mathbf{1}\right)\mathbf{=}\mathbf{4}\mathbf{;}\mathbf{y}\left(\mathbf{-}\mathbf{1}\right)\mathbf{=}\mathbf{-}\mathbf{3}$.

Short answer

Answer:

(a) $c=y^3+y+4x^2+5x$

(b)$$\begin{gathered} -2=y^3+y+4x^2+5x \\ 10=y^3+y+4x^2+5x \\ 67=y^3+y+4x^2+5x \\ -31=y^3+y+4x^2+5x \end{gathered}$$

Step-by-step solution

Integrate

Consider the differential equation$\frac{dy}{dx}=-\frac{8x+5}{3y^2+1}$

Separate the variables by cross multiplying$\left(3y^2+1\right)dy=-(8x+5)dx$

Now integrate to get:

$$\begin{gathered} \int \left(3y^2+1\right)dy=\int -(8x+5)dx \\ {y}^3+y=-\left(4x^2+5x\right)+c \end{gathered}$$

The family of solutions to the differential equation is c = y³+y+4x²+5x

Graph

The graph for varying values ofis shown below.

Values:$c=0,\pm 5,\pm 10,\pm 15,\pm 20$

Apply initial condition(b)

Substitute the initial condition$y\left(0\right)=-1$into the solution we get

$$\begin{gathered} \backslash \mathrm{begingathered}c=-1-1+0+0 \\ c=-2 \end{gathered}$$

The explicit solution is$-2=y^3+y+4x^2+5x$

Substitute the initial condition$y\left(0\right)=2$into the solution we get

$$\begin{gathered} c=(2{)}^3+\left(2\right)+4(0{)}^2+5\left(0\right) \\ c=8+2+0+0 \\ c=10 \end{gathered}$$

The explicit solution is role="math" $10=y^3+y+4x^2+5x$.

Apply initial condition

Substitute the initial condition$y(-1)=4$into the solution we get

$$\begin{gathered} c=(4{)}^3+\left(4\right)+4(-1{)}^2+5(-1) \\ c=64+4+4-5 \\ c=67 \end{gathered}$$

The explicit solution is$67=y^3+y+4x^2+5x$

If we substitute the initial condition $y(-1)=-3$into the solution we get

$$\begin{gathered} c=(-3{)}^3+(-3)+4(-1{)}^2+5(-1) \\ c=-27-3+4-5 \\ c=-31 \end{gathered}$$

The explicit solution is $-31=y^3+y+4x^2+5x$.

Graph

The graph for varying value ofis:

$\text{Order of}c\text{- values from bottom to top:}c=-31,-2,10,67.$