Higher Order Ordinary Differential Equations

Linear Differential Equations

$$\frac{\text{d}^ny}{\text{d}x^n}+ p_1(x)\frac{\text{d}^{n-1}y}{\text{d}x^{n-1}}+ \;...\; +p_n(x)y =q(x)$$

Based on $q(x)$, the above equation is categorized into $2$ types:

• Homogenous if $q(x)=0$
• Non-homogenous if $q(x)\not=0$

For 1st semester

Only linear, ordinary differential equations with constant coefficients are required.

They can be written as:

$$\frac{\text{d}^ny}{\text{d}x^n}+ a_1\frac{\text{d}^{n-1}y}{\text{d}x^{n-1}}+ \;...\; +a_ny =q(x)$$

Solution

The general solution of the equation is $y=y_p+y_c$. $$

Here

• $y_p$ - particular solution
• $y_c$ - complementary solution

Particular solution

Doesn’t exist for homogenous equations. For non-homogenous equations check steps section of 2nd order ODE.

Complementary solution

Solutions assuming $LHS=0$ (as in a homogenous equation). $$

$$y_c = \sum_{i=1}^{n}{c_i\,y_i}$$

Here

• $c_i$ - constant coefficients
• $y_i$ - a linearly-independent solution

Linearly dependent & independent

$n$-th order linear differential equations have n linearly independent solutions. $$

Two solutions of a differential equation $u,v$ are said to be linearly dependent, if there exists constants $c_1,c_2\;(\not=0)$ such that $c_1u(x)+c_2v(x)=0$.

Otherwise, the solutions are said to be linearly independent, which means:

$$\sum_{i=1}^{n}{c_iy_i}=0\implies \forall{c_i}=0$$

Linear differential operators with constant coefficients

WTF?

I don’t understand anything in this section.

Differential operator

Defined as:

$$\text{D}^i=\frac{\text{d}^i}{\text{d}x^i}\;;\;n\in\mathbb{Z}^+$$

The above equation can be written using the differential operator:

$$\text{D}^ny+ a_1\text{D}^{n-1}y+ \;...\; +a_ny =q(x)$$

Here if $y$ is factored out (how tf?):$$

$$(\text{D}^n+ a_1\text{D}^{n-1}+ \;...\; +a_n )y = P(D)y = q(x)$$

where $P(D)=(\text{D}^n+a_1\text{D}^{n-1}+\;...\;+a_n)$. $$

$P(D)$ is called a polynomial differential operator with constant coefficients.$$