Question

Find the general solution of the given differential equation. $$ y^{\mathrm{iv}}-5 y^{\prime \prime}+4 y=0 $$

Short answer

Question: Find the general solution of the given fourth-order linear homogenous differential equation with constant coefficients: \(y^{''''} - 5y^{''} + 4y = 0\). Answer: The general solution of the given differential equation is: $$ y(x) = c_1 e^x + c_2 e^{-x} + c_3 e^{2x} + c_4 e^{-2x}. $$ where \(c_1, c_2, c_3,\) and \(c_4\) are arbitrary constants determined by the initial conditions.

Step-by-step solution

Write down the characteristic equation from the given differential equation.

To do this, we replace the derivatives with their corresponding powers of the variable "r". The characteristic equation for the differential equation $$ y^{''''} - 5y^{''} + 4y = 0 $$ becomes $$ r^4 - 5r^2 + 4 = 0. $$

Solve for the roots of the characteristic equation.

The characteristic equation is a quadratic equation in terms of \(r^2\): $$ (r^2)^2 - 5(r^2) + 4 = 0. $$ Now, let \(a = r^2\). We have the quadratic equation $$ a^2 - 5a + 4 = 0. $$ The roots of this quadratic equation can be found by factoring: $$ (a-1)(a-4) = 0. $$ This gives us the roots \(a = 1\) and \(a = 4\). To find the corresponding roots of the original characteristic equation, we substitute back the variable "r", representing the original roots: \(r^2 = 1\) and \(r^2 = 4\). For \(r^2 = 1\), we have \(r = \pm 1\). For \(r^2 = 4\), we have \(r = \pm 2\). The roots of the characteristic equation are \(r_1 = 1, r_2 = -1, r_3 = 2,\) and \(r_4 = -2\).

Write down the general solution using the roots from Step 2.

Since we have found four distinct roots for the characteristic equation, the general solution of the given differential equation can be written as the linear combination of the exponential functions corresponding to the roots: $$ y(x) = c_1 e^{1x} + c_2 e^{-1x} + c_3 e^{2x} + c_4 e^{-2x}. $$ where \(c_1, c_2, c_3,\) and \(c_4\) are arbitrary constants determined by the initial conditions. So, the general solution of the given differential equation is: $$ y(x) = c_1 e^x + c_2 e^{-x} + c_3 e^{2x} + c_4 e^{-2x}. $$