Problem 30 - Advanced Engineering Mathematics

Question

Verify that the given differential equation is not exact. Multiply the given differential equation by the indicated integrating factor \(\mu(x, y)\) and verify that the new equation is exact. Solve. $$ \left(x^{2}+2 x y-y^{2}\right) d x+\left(y^{2}+2 x y-x^{2}\right) d y=0 ; \mu(x, y)=(x+y)^{-2} $$

Step-by-Step Solution

Answer

The solution is \(-\frac{x}{x+y} = C\).

Step 1: Check Exactness of the Original Equation

For a differential equation of the form \(M(x, y) \, dx + N(x, y) \, dy = 0\), it is exact if and only if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).In this exercise, \(M(x, y) = x^2 + 2xy - y^2\) and \(N(x, y) = y^2 + 2xy - x^2\).Calculate \(\frac{\partial M}{\partial y}\):\[\frac{\partial M}{\partial y} = 2x - 2y\]Calculate \(\frac{\partial N}{\partial x}\):\[\frac{\partial N}{\partial x} = 2y - 2x\]Since \(\frac{\partial M}{\partial y} e \frac{\partial N}{\partial x}\), the equation is not exact.

Step 2: Apply the Integrating Factor

The integrating factor \(\mu(x, y) = (x+y)^{-2}\) is given. Multiply the original differential equation by this factor:\[(x+y)^{-2}(x^{2}+2 x y-y^{2}) \, dx + (x+y)^{-2}(y^{2}+2 x y-x^{2}) \, dy = 0\]Let the new terms be \(\hat{M}(x, y)\) and \(\hat{N}(x, y)\):\[\hat{M}(x, y) = (x+y)^{-2}(x^{2}+2 x y-y^{2})\]\[\hat{N}(x, y) = (x+y)^{-2}(y^{2}+2 x y-x^{2})\]

Step 3: Verify Exactness of the New Equation

Compute \(\frac{\partial \hat{M}}{\partial y}\) and \(\frac{\partial \hat{N}}{\partial x}\):\[\frac{\partial \hat{M}}{\partial y} = (x+y)^{-2}(2x - 2y) + 2(x+y)^{-3}(x^2 + 2xy - y^2)\]\[\frac{\partial \hat{N}}{\partial x} = (x+y)^{-2}(2y - 2x) + 2(x+y)^{-3}(y^2 + 2xy - x^2)\]After simplification, \(\frac{\partial \hat{M}}{\partial y} = \frac{\partial \hat{N}}{\partial x}\) indicating that the new equation is exact.

Step 4: Solve the Exact Differential Equation

To solve, integrate \(\hat{M}(x, y)\) with respect to \(x\):\[\int (x+y)^{-2}(x^2 + 2xy - y^2) \, dx = -\frac{x}{x+y} + f(y)\]Differentiate the result with respect to \(y\) and set it equal to \(\hat{N}(x, y)\) to find \(f'(y)\):\[\frac{d}{dy}\left(-\frac{x}{x+y} + f(y)\right) = \frac{x}{(x+y)^2} + f'(y)\]Set this equal to \(\hat{N}(x, y) = \frac{y}{(x+y)^2}\) and solve:\[f'(y) = \frac{y}{(x+y)^2} - \frac{x}{(x+y)^2}\]\[f'(y) = 0\]Integrate to find \(f(y) = C\), where \(C\) is a constant. So, the solution is:\[-\frac{x}{x+y} = C\]

Step 5: Conclusion

The new differential equation, after multiplying by the integrating factor, is exact. Solving the exact equation yields the solution \(-\frac{x}{x+y} = C\).

Key Concepts

Integrating Factor

An integrating factor is a function, often denoted by \( \mu(x, y) \), used to make a non-exact differential equation exact. It acts like a magical multiplier that changes the entire equation so that its solution becomes straightforward. In many cases, the integrating factor depends on variables \(x\) and \(y\), as seen in our problem where \(\mu(x, y) = (x+y)^{-2}\).
To apply it correctly, follow these steps:

• Identify the original non-exact differential equation.
• Multiply each term in the equation by the integrating factor.
• Check the new equation for exactness (more on this in the "Exactness Check" section).

Using these steps, you convert a complex problem into a simpler one, making the equation ready to tackle with traditional solving techniques.

Partial Derivatives

Partial derivatives are fundamental when dealing with functions of two or more variables. They represent how a function changes as one of its variables is altered, keeping others constant. For a differential equation to be exact, we must calculate these derivatives:
- \( \frac{\partial M}{\partial y} \): This evaluates how the function \( M(x, y) \) changes as \( y \) changes.- \( \frac{\partial N}{\partial x} \): This shows the rate of change in \( N(x, y) \) with respect to changes in \( x \).In the context of our problem, computing these derivatives helps us determine whether \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). If they do not, as initially observed, it signals that the line of the differential equation is not exact. After multiplying by the integrating factor, however, recalculating these verifies that the modified equation is transformed into an exact one.

Solving Differential Equations

Solving differential equations involves finding a function or a set of functions that satisfy the given equation. In this exercise, after ensuring our equation is exact, we follow a structured method to find the solution.
Steps include:

• Integrate \( \hat{M}(x, y) \) with respect to \( x \), treating \( y \) as a constant.
• This integration yields a function plus an arbitrary function of \( y \), \( f(y) \).
• Differentiating this result with respect to \( y \) helps us find \( f'(y) \) by matching it to \( \hat{N}(x, y) \).

The integration constant or function \( f(y) \) is determined such that the derived solution satisfies the adjusted equation. This concise method leverages the transformed exactness to produce a straightforward solution.

Exactness Check

Exactness check is like a consistency test to see whether the work on a differential equation is correct and it can be solved easily. For the expression \( M(x, y) \, dx + N(x, y) \, dy = 0 \) to be exact, \( \frac{\partial M}{\partial y} \) must equal \( \frac{\partial N}{\partial x} \).
If this condition holds, the equation is exact and can be solved directly. As in our case, the original equation did not satisfy \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), indicating that it was not exact initially.
Post application of the integrating factor, a fresh check showed they are indeed equal, confirming exactness. This step is crucial because it sets the stage for applying integration techniques to find precise solutions.