Question

Question: (a) Find an implicit solution of the IVP

(b) Use part (a) to find an explicit solution$\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$of the IVP.

(c) Consider your answer to part (b) as a function only. Use a graphing utility or a CAS to graph this function, and then use the graph to estimate its domain.

(d) With the aid of a root-finding application of a CAS, determine the approximate largest interval I of definition of the solution$\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ in part (b). Use a graphing utility or a CAS to graph the solution curve for the IVP on this interval.

Short answer

Answer: Solution of part (a) is $y^2+2y=x^4+3x^2+3$. Solution of part (b) is $y=-1\pm \sqrt{x^4+3x^2+4}$. Solution of part (c) is $x\in (-\infty ,\infty )\text{.}$And the graph of the function is shown as below:-

Solution of part (d) is $(-\infty ,\infty )$.

Step-by-step solution

Definition of implicit function

Implicit function is a function with multiple variables, and one of the variables is a function of the other set of variables

Evaluation 

Consider $(2y+2)dy-\left(4x^3+6x\right)dx=0,y\left(0\right)=-3$

$(2y+2)dy-\left(4x^3+6x\right)dx=0$

This equation is separable so we have

$\int (2y+2)dy=\int \left(4x^3+6x\right)dx{y}^2+2y$

$=x^4+3x^2+c$

The form for the solution is implicit since the equation does not have explicitly.

Now we can solve for by plugging in the initial condition$y\left(0\right)=-3$

$$\begin{gathered} (-3{)}^2+2(-3)=0^4+3·0^2+c \\ 9-6=c \\ 3=c \end{gathered}$$

So the implicit solution to the initial value problem is $y^2+2y=x^4+3x^2+3$

(b)

To find the explicit solution to equation (1) by using the quadratic formula to solve for .

$$\begin{gathered} y^2+2y-\left(x^4+3x^2+3\right)=0 \\ y=\frac{-2\pm \sqrt{2^2-4·1·\left(-x^4-3x^2-3\right)}}{2·1} \\ y=\frac{-2\pm \sqrt{4\left(x^4+3x^2+4\right)}}{2} \\ y=-1\pm \sqrt{x^4+3x^2+4} \end{gathered}$$

(c)

The graph below shows the plots of the two equations given as

$$\begin{gathered} y=-1+\sqrt{x^4+3x^2+4} \\ y=-1-\sqrt{x^4+3x^2+4} \end{gathered}$$

Equation (1) is in red and equation (2) is in green.

.

Looking at the graph, the domain of the function is all real values of or in interval form:$x\in (-\infty ,\infty )\text{.}$

(d)

Taking into consideration the initial condition, $y\left(0\right)=-3$the solution to the initial value problem is then defined in equation (3) since that is the only equation that allows for y negative values. The graph is below. The largest interval for is $(-\infty ,\infty )$.