Question

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

Short answer

The particular solution of given differential equation is \(f(x) = \frac{1}{2}e^{x} + \frac{1}{2}e^{-x} + 1\).

Step-by-step solution

Identify the type of differential equation

The given differential equation is of the form \(f''(x) = g(x)\), where \(g(x) = \frac{1}{2}(e^x + e^{-x})\). This is a second order non-homogeneous differential equation.

Determine the complementary solution

The complementary solution, \(f_c(x)\), of the equation is the solution of the homogeneous form, i.e when \(g(x) = 0\). So, we solve the equation \(f''(x) = 0\). The general solution of this differential equation is \(f_c(x) = Ax + B\), where A and B are constants.

Find the particular solution

To solve for particular solution, we make a guess based on the right hand side of the equation. In this case, it is of the form \(Ce^{x} + De^{-x}\). Differentiating this guess once gives \(Ce^{x} - De^{-x}\) and differentiating again gives \(Ce^{x} + De^{-x}\). Equating this to our right hand side \(\frac{1}{2}(e^x + e^{-x})\), we get \(C = D = 1/2.\) Therefore, the particular solution, \(f_p(x)\), is \(\frac{1}{2}e^{x} + \frac{1}{2}e^{-x}\).

Form the general solution

The general solution is the sum of the complementary and the particular solution, giving us \(f(x) = Ax + B + \frac{1}{2}e^{x} + \frac{1}{2}e^{-x}\).

Apply the initial conditions to find A and B

Apply the initial conditions, \(f(0) = 1\) and \(f'(0) = 0\). After plugging these values into \(f(x)\) and \(f'(x)\) respectively, we obtain \(A = 0\) and \(B = 1\).