03 - Solving Second Order Homogenous ODEs

Auxiliary Equation

$$m^2+qm+r=0$$

Solving Second Order Homogenous ODEs

Consider the homogenous linear equation

$$\frac{d^{2}y}{d{x}^2}+q\frac{dy}{dx}+ry=0$$

where $q,r$ are real numbers

Then the Auxiliary Equation has two roots $m_1,m_2$

If $m_1\ne m_2$ are real then the general solution is

$$y(x)=C_1{e}^{m_{1}x}+C_2{e}^{m_{2}x}$$

If $m_1=m_2=m$ is a repeated real root then the general solution is

$$y(x)=(C_{1}x+C_2)e^{mx}$$

If $m_1=\alpha +i\beta$ is a complex root so that $m_2=\alpha -i\beta$ then the general solution is

$$y(x)=e^{\alpha x}(C_1\cos (\beta x)+C_2\sin (\beta x))$$

where $C_1,C_2$ are constants

Proof

We can rewrite the ODE as

$$\frac{d^{2}y}{d{x}^2}-(m_1+m_2)\frac{dy}{dx}+m_1{m}_{2}y=0$$

As $e^{m_{1}x}$ is a solution to this ODE (can check this by direct substitution)

So using we can try $y(x)=u(x)e^{m_{1}x}$ as a secondary solution to the ODE
Then

$$\frac{dy}{dx}=\left(\frac{du}{dx}+m_{1}u\right)e^{m_{1}x}\frac{d^{2}y}{d{x}^2}=\left(\frac{d^{2}u}{d{x}^2}+2m_1\frac{du}{dx}+m_1^{2}u\right)e^{m_{1}x}$$

Hence

$$\left(\frac{d^{2}u}{d{x}^2}+2m_1\frac{du}{dx}+m_1^{2}u\right)-(m_1+m_2)\left(\frac{du}{dx}+m_{1}u\right)+m_1{m}_{2}u=0$$

Then

$$\frac{d^{2}u}{d{x}^2}-(m_2-m_1)\frac{du}{dx}=0$$

Therefore if $m_1\ne m_2$ then

$$\frac{du}{dx}=k{e}^{(m_2-m_1)x}\text{ where }k\in \mathbb{R}$$

Hence

$$u(x)=C_2{e}^{(m_2-m_1)x}+C_1\text{ where }{C}_1,C_2\in \mathbb{R}$$

So we get the $m_1\ne m_2$ case

For $m_1=m_2$ then

$$\frac{d^{2}u}{d{x}^2}=0⟹u(x)=C_{1}x+C_2$$

So we get the $m_1=m_2$ case

For Case 3 using $e^{i\theta }=\cos \theta +i\sin \theta$

Applying it to higher order derivatives

Same concept applies as you just look at the roots and apply the same rules as before