Q16E

Question

Using the Wronskian in this Problem, verify that the given functions form a fundamental solution set for the given differential equation and find a general solution.

$y''' - y'' + 4 y' - 4 y = 0; {e^{x}, \cos 2 x, \sin 2 x}$

Step-by-Step Solution

Verified

Answer

Thus, it is verified that the given functions form a fundamental solution set for the given differential equation, and the general solution is $y = {Ae}^{x} + Bcos 2 x + Csin 2 x .$

1Step 1: Using the concept of Wronskian

The given function is $\{e^{x}, \cos 2 x, \sin 2 x\} .$

Apply the concept of Wronskian,

$W [f_{1}, f_{2}, \dots, f_{n}] = |\begin{matrix} f_{1} (x) & f_{2} (x) & \dots & f_{n} (x) \\ f_{1}' (x) & f_{2}' (x) & \dots & f_{n}' (x) \\ ⋮ & ⋮ & ⋮ \\ {f_{1}}^{n - 1} (x) & {f_{2}}^{n - 1} (x) & \dots & {f_{n}}^{n - 1} (x) \end{matrix}|$

Therefore,

$W [e^{x}, \cos 2 x, \sin 2 x] = |\begin{matrix} e^{x} & \cos 2 x & \sin 2 x \\ e^{x} & - 2 \sin 2 x & 2 \cos 2 x \\ e^{x} & - 4 \cos 2 x & - 4 \sin 2 x \end{matrix}|$

Solve the above equation,

$W [e^{x}, \cos 2 x, \sin 2 x] = |\begin{matrix} e^{x} & \cos 2 x & \sin 2 x \\ e^{x} & - 2 \sin 2 x & 2 \cos 2 x \\ e^{x} & - 4 \cos 2 x & - 4 \sin 2 x \end{matrix}| = e^{x} [8 {(\sin 2 x)}^{2} + 8 {(\cos 2 x)}^{2}] - \cos 2 x (- 4 e^{x} \sin 2 x - 2 e^{x} \cos 2 x) + \sin 2 x (- 4 e^{x} \cos 2 x + 2 e^{x} \sin 2 x) = 8 e^{x} {(\sin 2 x)}^{2} + 8 e^{x} {(\cos 2 x)}^{2} + 4 e^{x} \sin 2 x \cos 2 x + 2 e^{x} {(\cos 2 x)}^{2} - 4 e^{x} \sin 2 x \cos 2 x + 2 e^{x} {(\sin 2 x)}^{2} = 10 e^{x} {(\sin 2 x)}^{2} + 10 e^{x} {(\cos 2 x)}^{2} = 10 e^{x}$