Question

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

Short 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-by-step solution

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.

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.

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 \(E = B + C + D\).

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 \(A = \frac{1}{2} - 1\).

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 \(f(x) = -\frac{1}{2}x + \frac{7}{4} - \cos x + \frac{1}{2} e^{2x}\).