Question

Solve the differential equation. $$ f^{\prime \prime}(x)=\frac{2}{x^{2}}, f^{\prime}(1)=4, f(1)=3 $$

Short answer

The solution for the given differential equation is \(f(x)= -2\ln|x| + 6x -3\).

Step-by-step solution

Solve the Homogeneous Differential Equation

First, integrate the differential equation \(f^{\prime \prime}(x)=\frac{2}{x^{2}}\) with respect to \(x\). The result of the integral \(\int f^{\prime \prime}(x) dx\) is \(f^{\prime}(x)\). The integral of \(\frac{2}{x^{2}}\) with respect to \(x\) gives us \(-\frac{2}{x}\).

Apply the First Initial Condition

Apply the initial condition \(f^{\prime}(1)=4\), which gives us a constant \(C\) to our solution from step 1. Put \(x=1\) in the equation from step 1, \(f^{\prime}(1)= -\frac{2}{1} + C\), we can calculate the constant as \(C=4+2=6\). Therefore, \(f^{\prime}(x)= -\frac{2}{x} + 6\).

Solve the Non-homogeneous Differential Equation

Integrate the result \(f^{\prime}(x)= -\frac{2}{x} + 6\) with respect to \(x\) to solve for \(f(x)\). The integral of \(-\frac{2}{x}\) with respect to \(x\) gives us \(-2\ln|x|\). The integral of \(6\) with respect to \(x\) gives us \(6x\). Thus, the result of this integral is \(f(x)=-2\ln|x| + 6x\).

Apply Second Initial Condition

Using the second boundary condition \(f(1)=3\), calculate the second constant \(C\). Substitute \(x=1\) in the equation \(f(x)=-2\ln|1| + 6\cdot 1 + C\), this gives us \(C=3-6=-3\). Therefore, the solution for the differential equation is \(f(x)= -2\ln|x| + 6x -3\).