Problem 51 - Advanced Engineering Mathematics

Question

(a) Find an implicit solution of the IVP $$ (2 y+2) d y-\left(4 x^{3}+6 x\right) d x=0, \quad y(0)=-3 $$ (b) Use part (a) to find an explicit solution \(y=\phi(x)\) 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 \(y=\phi(x)\) in part (b). Use a graphing utility or a CAS to graph the solution curve for the IVP on this interval.

Step-by-Step Solution

Answer

(a) Implicit solution: \(y^2 + 2y = x^4 + 3x^2 + 3\). (b) Explicit solution: \(y = -1 - \sqrt{x^4 + 3x^2 + 4}\). Domain is all real numbers.

Step 1: Write down the given equation

The initial value problem is given by \((2y+2)dy - (4x^3 + 6x)dx = 0\) with the initial condition \(y(0) = -3\). This is a differential equation in differential form.

Step 2: Express the problem in standard form

Rearrange the terms to obtain: \((2y+2)dy = (4x^3 + 6x)dx\). This can be rewritten as \( \frac{dy}{dx} = \frac{4x^3 + 6x}{2y + 2} \).

Step 3: Integrate both sides

To find an implicit solution, integrate both sides: \( \int (2y+2) dy = \int (4x^3 + 6x) dx \). The integrals result in \( y^2 + 2y = x^4 + 3x^2 + C \).

Step 4: Apply the initial condition

Use the initial condition \(y(0) = -3\): \((-3)^2 + 2(-3) = 0^4 + 3(0)^2 + C\). This gives \(9 - 6 = C\), thus \(C = 3\).

Step 5: Write the implicit solution

The implicit solution found is \(y^2 + 2y = x^4 + 3x^2 + 3\).

Step 6: Solve for y explicitly

From the implicit solution \(y^2 + 2y = x^4 + 3x^2 + 3\), we complete the square to find \(y = -1 \pm \sqrt{x^4 + 3x^2 + 4}\). Since \(y(0) = -3\), the negative sign is taken: \(y = -1 - \sqrt{x^4 + 3x^2 + 4}\).

Step 7: Graph the explicit solution

Graph \(y = -1 - \sqrt{x^4 + 3x^2 + 4}\) using a graphing utility. The function is real only when the expression under the square root is non-negative, constraining \(x^4 + 3x^2 + 4 \geq 0\).

Step 8: Estimate the domain from the graph

Since the expression inside the square root is always positive, the domain of \(y = \phi(x)\) is all real numbers \(\mathbb{R}\).

Step 9: Determine interval of definition

Considering any root-finding, the polynomial inside the square root has no real roots that influence the expression being real, so the interval of definition spans all real numbers.

Key Concepts

Initial Value Problem (IVP)

An Initial Value Problem (IVP) is an important concept in differential equations, where you need to find a solution to a differential equation that meets a specific initial condition. Initial conditions are specific values given for the function at a particular point. For example, in the problem, the initial condition is given as \( y(0) = -3 \). This means that when \( x = 0 \), the value of the function \( y \) is \( -3 \). This condition helps identify a unique solution from potentially many solutions that satisfy the differential equation.IVPs are solved using the initial condition as a boundary constraint, which is essential to find a specific solution tailored to the situation described. By knowing the initial state, you can integrate the differential equation and use the initial condition to solve for any constants that arise from the integration process.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In the problem provided, we are given a differential equation in the differential form: \( (2y+2)dy - (4x^3 + 6x)dx = 0 \). This represents a relationship between the variables \( x \) and \( y \) and their rates of change.There are different types of differential equations: ordinary, partial, linear, and nonlinear. The equation we are dealing with is an ordinary differential equation (ODE) as it involves ordinary derivatives (not partial derivatives). By transforming this differential form equation into the standard form, \( \frac{dy}{dx} = \frac{4x^3 + 6x}{2y + 2} \), it becomes an explicit expression of the rate of change of \( y \) with respect to \( x \). This makes it easier to see the relationship and proceed with finding its solution through integration.

Domain of Solutions

The domain of solutions refers to the set of all possible inputs (in this case, values of \( x \)) for which the differential equation retains validity and the solution makes sense. For the explicit solution \( y = -1 - \sqrt{x^4 + 3x^2 + 4} \), the domain is estimated through examining the conditions under which the expression is real.Since the expression under the square root, \( x^4 + 3x^2 + 4 \), must remain non-negative, we need to verify when this condition holds. Through graphing tools or algebraic analysis, we observe that \( x^4 + 3x^2 + 4 \) is always greater than zero for all real \( x \). Therefore, the domain of \( y = \phi(x) \) is all real numbers, \( \mathbb{R} \). This means the solution is defined and valid for any real input value of \( x \).

Integration Methods

Integration is a critical process in finding the solutions to differential equations. In this task, we employ indefinite integration on both sides of the transformed differential equation \( \int (2y+2) dy = \int (4x^3 + 6x) dx \). By integrating, we find antiderivatives that represent a general solution to the equation.For the given problem, integrating the left side \( \int (2y+2) dy \) gives \( y^2 + 2y + C_1 \), and integrating the right side \( \int (4x^3 + 6x) dx \) gives \( x^4 + 3x^2 + C_2 \). By equating these results and applying the initial condition, we solve for constants which are unknown otherwise.Integration methods can include direct integration, substitution, integration by parts, and more, depending on the complexity of the function involved. Here, direct integration was straightforward since the polynomial terms were simple to handle, allowing us to find the implicit and eventually the explicit solution.