Problem 49 - Advanced Engineering Mathematics

Question

(a) An implicit solution of the differential equation \((2 y+2) d y-\left(4 x^{3}+6 x\right) d x=0\) is \\[y^{2}+2 y-x^{4}-3 x^{2}+c=0\\]. The condition \(y(0)=-3\) implies that \(c=-3 .\) Therefore \(y^{2}+2 y-x^{4}-3 x^{2}-3=0\). (b) Using the quadratic formula we can solve for \(y\) in terms of \(x\): \\[y=\frac{-2 \pm \sqrt{4+4\left(x^{4}+3 x^{2}+3\right)}}{2}\\]. The explicit solution that satisfies the initial condition is then \\[y=-1-\sqrt{x^{4}+3 x^{3}+4}\\]. (c) From the graph of the function \(f(x)=x^{4}+3 x^{3}+4\) below we see that \(f(x) \leq 0\) on the approximate interval \(-2.8 \leq x \leq-1.3 .\) Thus the approximate domain of the function $$y=-1-\sqrt{x^{4}+3 x^{3}+4}=-1-\sqrt{f(x)}$$ is \(x \leq-2.8\) or \(x \geq-1.3 .\) The graph of this function is shown below. (d) Using the root finding capabilities of a CAS, the zeros of \(f\) are found to be -2.82202 and \(-1.3409 .\) The domain of definition of the solution \(y(x)\) is then \(x>-1.3409 .\) The equality has been removed since the derivative \(d y / d x\) does not exist at the points where \(f(x)=0 .\) The graph of the solution \(y=\phi(x)\) is given on the right.

Step-by-Step Solution

Answer

The function \( y(x) = -1 - \sqrt{x^4 + 3x^2 + 3} \) is defined for \( x > -1.3409 \).

Step 1: Identify Implicit Solution and Initial Condition

The given implicit solution for the differential equation is \( y^2 + 2y - x^4 - 3x^2 + c = 0 \). We use the initial condition \( y(0) = -3 \) to solve for \( c \). Substitute \( x = 0 \) and \( y = -3 \) into the equation:\[(-3)^2 + 2(-3) - 0^4 - 3\cdot0^2 + c = 0 \9 - 6 + c = 0 \c = -3\]Thus, the equation becomes \( y^2 + 2y - x^4 - 3x^2 - 3 = 0 \).

Step 2: Solve Implicit Equation Using Quadratic Formula

To make the solution explicit by solving for \( y \), use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for the equation:\[y^2 + 2y - (x^4 + 3x^2 + 3) = 0\]where \( a = 1 \), \( b = 2 \), and \( c = -(x^4 + 3x^2 + 3) \). This gives:\[y = \frac{-2 \pm \sqrt{4 + 4(x^4 + 3x^2 + 3)}}{2} \y = -1 \pm \sqrt{(x^4 + 3x^2 + 3)}\]

Step 3: Select the Explicit Solution with Initial Condition

Given the initial condition \( y(0) = -3 \), we choose the negative part of the solution to satisfy this:\[ y = -1 - \sqrt{x^4 + 3x^2 + 3} \]This matches \( y = -3 \) when \( x = 0 \).

Step 4: Determine Domain of Solution

To find the domain, examine when \( y = -1 - \sqrt{x^4 + 3x^2 + 3} \) is defined. It requires\[ f(x) = x^4 + 3x^2 + 3 \geq 0 \ \text{for } \sqrt{f(x)} ext{ to be real}.\]Using a root-finding approach, we determine \( f(x) \geq 0 \) for \( x > -1.3409 \) based on the roots \( x = -2.82202 \) and \( x = -1.3409 \).

Key Concepts

Implicit Solution

In differential equations, an implicit solution is a way of expressing the relationship between variables without explicitly solving for one variable in terms of the other(s). Unlike explicit solutions, which provide a direct formula for one variable, implicit solutions are often found as equations involving multiple variables.

• For example, the implicit solution given in the original exercise is expressed as \( y^2 + 2y - x^4 - 3x^2 + c = 0 \).
• This equation does not solve explicitly for \( y \) in terms of \( x \), but it still satisfies the differential equation.

To use implicit solutions effectively, especially in complex equations, you may need initial conditions to find specific constants, as demonstrated when solving for \( c \) using \( y(0) = -3 \). This approach preserves the relationship among all variables without isolating one upfront, making it versatile in various applications.

Initial Condition

Initial conditions are used to find specific solutions to differential equations, which often have multiple general solutions. An initial condition like \( y(0) = -3 \) provides a specific value for the dependent variable at a given point and helps determine any unknown constants in an implicit solution.

• By plugging the initial condition into the implicit equation, specific values can be determined to make sure the solution matches the scenario being examined.
• In our exercise, using the initial condition led us to determine that \( c = -3 \).

These conditions are crucial in applications where only one solution fits the behavior of a system or process, such as in physics or engineering studies. They make the substantial difference between a general equation and a solution aligned with real-world data.

Quadratic Formula

The quadratic formula is a fundamental tool in algebraic problem-solving, allowing us to find solutions for quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula is especially useful when solving implicit equations to find explicit solutions.For the equation \( y^2 + 2y - (x^4 + 3x^2 + 3) = 0 \), we use the quadratic formula:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = 2 \), and \( c = -(x^4 + 3x^2 + 3) \). This gives:

• Solving with the plus or minus sign shows two potential solutions, \( y = -1 \pm \sqrt{x^4 + 3x^2 + 3} \).

In practice, knowing whether to choose the positive or negative root often relies on other conditions, like initial conditions, to fit the behavior of the function more precisely.

Domain of Definition

The domain of definition for a function is the set of all possible input values (often \( x \) values) for which the function is defined. In our case, for the explicit solution \( y = -1 - \sqrt{x^4 + 3x^2 + 3} \), it is crucial to determine when the expression under the square root is non-negative, since we cannot take the square root of a negative number in real numbers. Graphical and computational methods often assist in this process.

• Using a CAS, the zeros of the function \( f(x) = x^4 + 3x^2 + 3 \) were found to be approximately \( -2.82202 \) and \( -1.3409 \).
• This implies that the function \( y \) is defined for \( x > -1.3409 \), after checking that the derivative \( dy/dx \) does not exist at the points where \( f(x) = 0 \).

A precise understanding of the domain ensures that any real-valued function based on the expression is defined appropriately, preventing mathematical and logical errors in application.