Problem 15 - Advanced Engineering Mathematics

Question

In Problems 9-16, solve the given differential equation. $$ \left(x^{2}+4\right) d y=(2 x-8 x y) d x $$

Step-by-Step Solution

Answer

Separate variables, integrate, and solve for \(y\). Solution: \(y = \frac{1}{4} (1 - K (x^2 + 4)^{-4})\).

Step 1: Separate Variables

The given differential equation is \((x^2 + 4)\, dy = (2x - 8xy)\, dx\). Our goal is to separate the variables such that terms involving \(y\) are on one side and terms involving \(x\) are on the other side. Divide both sides by \((x^2 + 4)\) and \((1 - 4y)\):\[\frac{dy}{1 - 4y} = \frac{2x}{x^2 + 4} dx.\]

Step 2: Integrate Both Sides

Now, integrate both sides to find the general solution. Integrate the left side with respect to \(y\) and the right side with respect to \(x\):\[\int \frac{1}{1 - 4y} dy = \int \frac{2x}{x^2 + 4} dx.\]

Step 3: Integrate the Left Side

For the left side, the integral is:\[-\frac{1}{4} \ln|1 - 4y| + C_1.\]

Step 4: Integrate the Right Side

For the right side, use a substitution method. Let \(u = x^2 + 4\), then \(du = 2x dx\). Thus, the integral becomes:\[ \int \frac{du}{u} = \ln|u| + C_2.\]Substitute back the original variable:\[ \ln|x^2 + 4| + C_2.\]

Step 5: Combine Integrals

The equation from integrating both sides becomes:\[-\frac{1}{4} \ln|1 - 4y| = \ln|x^2 + 4| + C.\]Where \(C = C_2 - C_1\).

Step 6: Solve for the General Solution if Possible

To find \(y\) in terms of \(x\), solve the equation\[-\frac{1}{4} \ln|1 - 4y| = \ln|x^2 + 4| + C.\]Exponentiate both sides to eliminate the logarithm:\[ |1 - 4y|^{1/4} = rac{1}{(x^2 + 4) e^C}.\]

Step 7: Simplify the General Solution

Solving for \(y\), we get:\[1 - 4y = (x^2 + 4)^{-4} e^{-4C} = K (x^2 + 4)^{-4},\] where \(K = e^{-4C}\).Thus, the solution is\[y = \frac{1}{4} (1 - K (x^2 + 4)^{-4}).\]

Key Concepts

Variable Separation

Differential equations can often be solved by the technique of "variable separation". This involves rearranging the equation so that all terms containing one variable appear on one side of the equation and all terms containing the other variable appear on the opposite side. In the given equation

• The terms involving the variable \(y\) and \(dy\) must be separated from those involving \(x\) and \(dx\).
• To do this, divide both sides of the equation by any factors necessary to isolate \(dy\) and \(dx\).
• This results in a format where integration can occur with respect to each variable independently.

In our example: \[\frac{dy}{1 - 4y} = \frac{2x}{x^2 + 4} dx.\]This equation effectively separates the two variables, allowing the next step of integration to be tackled more easily.

Integration Techniques

When you have a differential equation that has been separated into distinct variable groups, the next step is integrating both sides. Integration techniques can vary based on the complexity of each side:

• Left Side: For integrals like \(\int \frac{1}{1 - 4y} \, dy\), we recognize a standard form that directly relates to the natural logarithm, giving us: \[-\frac{1}{4} \ln |1 - 4y| + C_1.\]
• Right Side: For \(\int \frac{2x}{x^2 + 4} \, dx\), we use substitution. Here, set \(u = x^2 + 4\), resulting in \( du = 2x \, dx\). This transforms the integral into: \[\int \frac{du}{u} = \ln |u| + C_2.\]

These methods effectively simplify complex expressions into more manageable functions that can be integrated with ease. Recognizing common forms or using substitutions is crucial to solving integrals.

General Solution Finding

After integrating both sides, the task is to solve for \(y\) to get the general solution of the differential equation. Begin with combining the results from both integrals:\[-\frac{1}{4} \ln |1 - 4y| = \ln |x^2 + 4| + C.\]The constant \(C\) represents an integration constant that can encapsulate any difference between \(C_1\) and \(C_2\). Then, to remove the logarithmic terms, exponentiate both sides:

• This yields: \(|1 - 4y|^{1/4} = \frac{1}{(x^2 + 4) e^C}.\)
• Solving for \(y\), you find:\[y = \frac{1}{4}(1 - K(x^2 + 4)^{-4}),\]
• where \(K = e^{-4C}\) represents a constant derived from the integration process.

This final expression is your general solution, expressing \(y\) in a function of \(x\) and encompassing all particular solutions.