Problem 62 - Essential Calculus

Question

Find the particular solution of the differential equation that satisfies the initial conditions. $$ f^{\prime \prime}(x)=\sin x+e^{2 x}, \quad f(0)=\frac{1}{4}, f^{\prime}(0)=\frac{1}{2} $$

Step-by-Step Solution

Answer

The particular solution of the given differential equation that satisfies the initial conditions is \(f(x) = -\frac{1}{2}x + \frac{7}{4} - \cos x + \frac{1}{2} e^{2x}\).

Step 1: Solve the Homogeneous Differential Equation

The given differential equation is not homogeneous, and it is noted by observing the structure of the equation \(f''(x) = \sin x + e^{2x}\). To solve the homogeneous equation, we would have to solve \(f''(x) = 0\). This leads to general solution: \(f(x) = Ax + B\), where A and B are constants. However, this is not enough, we must further solve the inhomogeneous part.

Step 2: Solve the Inhomogeneous Differential Equation

The solutions of the non-homogeneous part of the differential equation can be found by integration. For \(\int \sin x dx\), the antiderivative is \(- \cos x + C\), and for \(\int e^{2x} dx \), the antiderivative is \(\frac{1}{2}e^{2x} + D\). So, the solutions of the non-homogeneous part are \(- \cos x + C\) and \(\frac{1}{2} e^{2x} + D\) respectively.

Step 3: Construct the Particular Solution

The particular solution of the equation is a sum of the solution of the homogeneous part and the non-homogeneous parts, so we have \(f(x) = Ax + B - \cos x + C + \frac{1}{2} e^{2x} + D\). By grouping the constants together we can rewrite this as \(f(x) = Ax + E - \cos x + \frac{1}{2} e^{2x}\), where \(E = B + C + D\).

Step 4: Apply the Initial Conditions

We are given \(f(0)= \frac{1}{4}\) and \(f'(0)=\frac{1}{2}\). We apply these initial conditions to our particular function and its derivative. Applying \(f(0)= \frac{1}{4}\) gives \(\frac{1}{4} = E - 1 + \frac{1}{2}\), which can be simplified to \(E = \frac{3}{4} + 1\).Applying \(f'(0) = \frac{1}{2}\) gives \(\frac{1}{2} = A + 0 + e^{0}\), which simplifies to \(A = \frac{1}{2} - 1\).

Step 5: Construct the Final Particular Solution

We replace \(E\) and \(A\) in the general equation with the values found from the initial conditions: \(E = \frac{7}{4}\) and \(A = -\frac{1}{2}\). This gives the final particular solution of \(f(x) = -\frac{1}{2}x + \frac{7}{4} - \cos x + \frac{1}{2} e^{2x}\).

Key Concepts

Particular Solution

In the world of differential equations, finding the particular solution is often like solving a puzzle. It involves combining the solutions of both homogeneous and inhomogeneous parts of the equation. For our specific problem, the equation is noted as \( f''(x) = \sin x + e^{2x} \). The particular solution is designed to satisfy this entire equation, including any additional functions affecting it.

Here is how it works:

• First, we find the general solution for the homogeneous part (i.e., where the right-hand side equals zero). For the equation \( f''(x) = 0 \), a standard form, the solution is \( f(x) = Ax + B \), where \( A \) and \( B \) are constants.
• Next, we handle the inhomogeneous parts separately. Each term on the right needs an individual solution. In this case, the term \( \sin x \) has an antiderivative of \( -\cos x + C \), and \( e^{2x} \) becomes \( \frac{1}{2}e^{2x} + D \) when integrated.
• Finally, combine these results into a single expression. The particular solution includes both homogeneous and inhomogeneous results: \( f(x) = Ax + E - \cos x + \frac{1}{2}e^{2x} \), where \( E = B + C + D \).

By artfully assembling these pieces, the particular solution includes all characteristics of the given differential equation.

Initial Conditions

Initial conditions are crucial as they help determine the unique solution among infinite possibles. The differential equation itself shows a general form, but initial conditions pin down specific values. For example, with \( f(0) = \frac{1}{4} \) and \( f'(0) = \frac{1}{2} \), these conditions tell us exactly where the curve intersects the axes.

Applying initial conditions:

• Plug in the given value to find \( E \) in \( f(0) = \frac{1}{4} = E - 1 + \frac{1}{2} \). Simplifying gives \( E = \frac{7}{4} \).
• Apply the derivative condition with \( f'(0) = \frac{1}{2} \) into \( f'(x) = A - \sin x + e^{2x} \). Substituting gives \( \frac{1}{2} = A + 0 + 1 \), leading us to find \( A = -\frac{1}{2} \).

By using these steps, we tailor the particular solution so it aligns exactly with the conditions set by the problem, ensuring the solution is not just theoretically correct but practically applicable.

Inhomogeneous Equation

Inhomogeneous equations are a standard part of differential equations, often acting as a bridge to many real-world phenomena. When you encounter a differential equation that includes terms like \( \sin x \) or \( e^{2x} \) added to its derivative terms, you're dealing with an inhomogeneous equation.

Handling inhomogeneous equations:

• The presence of \( \sin x + e^{2x} \) in \( f''(x) = \sin x + e^{2x} \) indicates that our approach needs to specifically address these non-zero components on the equation's right-hand side.
• Unlike a homogeneous equation, which can be solved by setting the right side to zero, the inhomogeneous equation requires you to find particular solutions for each additional function separately.
• Essentially, you solve for a part derived from non-derivative-dependent terms, integrating them one by one to find \( -\cos x + C \) and \( \frac{1}{2}e^{2x} + D \)

Being inhomogeneous means the solutions have to capture the behavior and influences of external functions, which makes the process a bit more nuanced and precisely applicable to a broad range of disciplines.