Problem 39 - Advanced Engineering Mathematics

Question

(a) Show that a one-parameter family of solutions of the equation $$ \left(4 x y+3 x^{2}\right) d x+\left(2 y+2 x^{2}\right) d y=0 $$ is \(x^{3}+2 x^{2} y+y^{2}=c\). (b) Show that the initial conditions \(y(0)=-2\) and \(y(1)=1\) determine the same implicit solution. (c) Find explicit solutions \(y_{1}(x)\) and \(y_{2}(x)\) of the differential equation in part (a) such that \(y_{1}(0)=-2\) and \(y_{2}(1)=1\). Use a graphing utility to graph \(y_{1}(x)\) and \(y_{2}(x)\).

Step-by-Step Solution

Answer

The one-parameter family is a solution; both initial conditions lead to \( c = 4 \). The explicit solutions are \( y_1(x) = -x^2 - 2 \) and \( y_2(x) = -x^2 + 2 \).

Step 1: Verify if the given family is a solution

To verify that the family of solutions \( x^3 + 2x^2y + y^2 = c \) is actually a solution of \[ (4xy + 3x^2) dx + (2y + 2x^2) dy = 0, \] let's implicitly differentiate the family with respect to \(x\):First, differentiate:\( 3x^2 + 4xy + 2x^2y' + 2yy' = 0. \)By solving for \( dy/dx \) (\(y'\)) we can check our original equationto find consistency:\( y' = -\frac{3x^2 + 4xy}{2x^2 + 2y}. \)Substitute \( y' \) back into the given differential equation and verify that both sides equal zero.This confirms the family as a solution.

Step 2: Validate the initial conditions

Substitute \( x = 0 \) and \( y = -2 \) into the solution \( x^3 + 2x^2y + y^2 = c \):\( 0 + 0 + (-2)^2 = c \rightarrow c = 4. \)Now substitute \( x = 1 \) and \( y = 1 \) into the solution:\( 1 + 2 \cdot 1^2 \cdot 1 + 1^2 = c \rightarrow c = 4. \)Thus, the initial conditions indeed determine the same implicit solution, \( c = 4 \).

Step 3: Find explicit solutions for given conditions

Given \( x^3 + 2x^2y + y^2 = 4 \), solve for \( y \) to get the explicit solutions.Use the quadratic formula to solve for \( y \):\[y = \frac{-2x^2 \pm \sqrt{(2x^2)^2 - 4(x^3 - 4)}}{2}.\]Simplify the expression to find \( y_1(x) \) and \( y_2(x) \).- For \( y_1(0) = -2 \), choose the root that gives \( y_1(x) = -x^2 - 2. \)- For \( y_2(1) = 1 \), solve: \( 1^3 + 2\cdot1^2\cdot y_2 + y_2^2 = 4 \) to find \( y_2(x) = -x^2 + 2. \)

Step 4: Graph the solutions

Using a graphing utility, plot \( y_1(x) = -x^2 - 2 \) and \( y_2(x) = -x^2 + 2 \).These plots will show the explicit solution behavior given each initial condition.- \( y_1(x) \) should pass through (0, -2)- \( y_2(x) \) should pass through (1, 1)Ensure the qualitative nature of the graphs align with the implicit solution boundary \( x^3 + 2x^2y + y^2 = 4. \)

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when a function is not given explicitly. Instead, it's presented as a relationship between variables. For instance, when you have an equation like \(x^3 + 2x^2y + y^2 = c\), neither the function \(y(x)\) nor \(x(y)\) is isolated. To differentiate such expressions, you need to take the derivative of both sides with respect to a particular variable, usually \(x\). This involves:

• Applying the usual rules of differentiation (power, product, chain rules, etc.).
• Treating \(y\) as a function of \(x\). Therefore, whenever you differentiate \(y\), attach \(y'\) (or \(\frac{dy}{dx}\)).

Then, solve the resulting equation for \(y'\) to understand how \(y\) changes with \(x\). Implicit differentiation often uncovers a lot about the relationship between variables hidden in complex equations.

Explicit Solutions

An explicit solution expresses one variable directly in terms of the other. In the context of differential equations, finding \(y\) as a function of \(x\) is considered finding an explicit solution. Given the implicit general solution \(x^3 + 2x^2y + y^2 = c\), we were able to formulate this into explicit solutions through algebraic manipulation and the quadratic formula. Here are the steps that we followed:

• Recognize that the equation is quadratic in terms of \(y\).


• Use the quadratic formula to solve for \(y\): \[y = \frac{-2x^2 \pm \sqrt{(2x^2)^2 - 4(x^3 - 4)}}{2}.\]


• Simplify the expression to get specific solutions based on initial conditions.

Each explicit solution, \(y_1(x)\) and \(y_2(x)\), is useful for representing different scenarios according to initial values.

Initial Value Problem

An initial value problem is a type of differential equation with a set condition that defines the solution's constants. This makes differential equations useful in real-world scenarios. In tasks like validating initial conditions, specific values of \(x\) and \(y\) are plugged into the solution to find a particular constant \(c\). For our problem, verification was made as follows:

• Substitute the initial values \(x = 0, y = -2\) into our implicit solution: the calculation confirmed that \(c = 4\).


• Similarly, \(x = 1, y = 1\) also gave us a consistent \(c = 4\).

This consistency confirms that these conditions set the specific solution within the general family's boundary.

Graphing Solutions

Graphing the solutions of a differential equation can be a powerful tool for understanding the behavior of the function over a range. It's even more important in illustrating both explicit and implicit solutions. For our problem:

• We got two explicit solutions: \(y_1(x) = -x^2 - 2\) and \(y_2(x) = -x^2 + 2\).


• Using graphing utilities, these are plotted to make visible their paths.


• Each graph passes through specific initial conditions: \(y_1(x)\) passes through \( (0, -2)\) and \(y_2(x)\) passes through \( (1, 1)\).

The graphing process not only validates the algebraic work but also provides an intuitive visual confirmation. Such visual analysis is crucial in cases where solutions are complex and diverse.