# What is the general solution of the differential equation? : (d^4y)/(dx^4) - 4 y = 0

May 12, 2018

$y = A {e}^{\sqrt{2} x} + B {e}^{- \sqrt{2} x} + C \cos \left(\sqrt{2} x\right) + D \sin \left(\sqrt{2} x\right)$

#### Explanation:

With linear operator $D = \frac{d}{\mathrm{dx}}$, this is:

$\left({D}^{4} - 4\right) y = 0$

$\left({D}^{2} - 2\right) \left({D}^{2} + 2\right) y = 0$

$\left(D - \sqrt{2}\right) \left(D + \sqrt{2}\right) \left(D - i \sqrt{2}\right) \left(D + i \sqrt{2}\right) y = 0$

Each of these factors has a simple solution, eg:

• $\left(D - \sqrt{2}\right) y = 0 \implies y ' = \sqrt{2} y \implies y = \alpha {e}^{\sqrt{2} x}$

• $\left(D - i \sqrt{2}\right) y = 0 \implies y ' = i \sqrt{2} y \implies y = \beta {e}^{i \sqrt{2} x}$

The complete solution is the superposition of individual solutions:

$\implies y \left(x\right) = A {e}^{\sqrt{2} x} + B {e}^{- \sqrt{2} x} + C {e}^{i \sqrt{2} x} + D {e}^{- i \sqrt{2} x}$

$C {e}^{i \sqrt{2} x} + D {e}^{- i \sqrt{2} x}$ can also be re-written using the Euler formula , arriving at:

$y = A {e}^{\sqrt{2} x} + B {e}^{- \sqrt{2} x} + C ' \cos \left(\sqrt{2} x\right) + D ' \sin \left(\sqrt{2} x\right)$

May 12, 2018

$y = A {e}^{\sqrt{2} x} + B {e}^{- \sqrt{2} x} + C \cos \left(\sqrt{2} x\right) + D \sin \left(\sqrt{2} x\right)$

#### Explanation:

We have:

$\frac{{d}^{4} y}{{\mathrm{dx}}^{4}} - 4 y = 0$ ..... [A]

This is a fourth order linear Homogeneous Differentiation Equation with constant coefficients. The standard approach is to find a solution, ${y}_{c}$ of the homogeneous equation by looking at the Auxiliary Equation, which is the polynomial equation with the coefficients of the derivatives.

Complementary Function

The Auxiliary equation associated with [A] is:

${m}^{4} - 4 = 0$

The challenge with higher order Differential Equation is solving the associated higher order Auxiliary equation. However, in this case we have a difference of two squares, thus we can immediately factorise the equation using ${A}^{2} - {B}^{2} \equiv \left(A - b\right) \left(A + B\right)$, so we get::

$\left({m}^{2} - 2\right) \left({m}^{2} + 2\right) = 0$

So we have:

${m}^{2} - 2 = 0 = > m = \pm \sqrt{2}$ (real and distinct)
${m}^{2} + 2 = 0 = > m = \pm \sqrt{2} i$ (pure imaginary)

The roots of the auxiliary equation determine parts of the solution, which if linearly independent then the superposition of the solutions form the full general solution.

• Real distinct roots $m = \alpha , \beta , \ldots$ will yield linearly independent solutions of the form ${y}_{1} = A {e}^{\alpha x}$, ${y}_{2} = B {e}^{\beta x}$, ...
• Real repeated roots $m = \alpha$, will yield a solution of the form $y = \left(A x + B\right) {e}^{\alpha x}$ where the polynomial has the same degree as the repeat.
• Complex roots (which must occur as conjugate pairs) $m = p \pm q i$ will yield a pairs linearly independent solutions of the form $y = {e}^{p x} \left(A \cos \left(q x\right) + B \sin \left(q x\right)\right)$

Thus the solution of the homogeneous equation [A] is:

$y = A {e}^{\sqrt{2} x} + B {e}^{- \sqrt{2} x} + {e}^{0 x} \left(C \cos \left(\sqrt{2} x\right) + D \sin \left(\sqrt{2} x\right)\right)$

$\setminus \setminus = A {e}^{\sqrt{2} x} + B {e}^{- \sqrt{2} x} + C \cos \left(\sqrt{2} x\right) + D \sin \left(\sqrt{2} x\right)$

Note this solution has $4$ constants of integration and $4$ linearly independent solutions, hence by the Existence and Uniqueness Theorem their superposition is the General Solution