03. Second Order Linear Differential Equations

3.1 Two Theorems

01 - Space of Solutions of a Homogenous Linear ODE is a Vector Space

Space of Solutions of an ODE is a a Vector Space

Let $y_1,y_2$ be solutions of a homogenous linear ODE
Let ${\alpha }_1,{\alpha }_2$ be real numbers

Then

$${\alpha }_1{y}_1+{\alpha }_2{y}_2\text{ is also a solution of the ODE}$$

Hence the solutions to the ODE form a Real Vector Space

Proof

We know that

$$a_k(x)\frac{d^k{y}_1}{d{x}^k}+a_{k-1}(x)\frac{d^{k-1}{y}_1}{d{x}^{k-1}}+\cdots +a_1(x)\frac{d{y}_1}{dx}+a_0(x)y_1=0$$ $$a_k(x)\frac{d^k{y}_2}{d{x}^k}+a_{k-1}(x)\frac{d^{k-1}{y}_2}{d{x}^{k-1}}+\cdots +a_1(x)\frac{d{y}_2}{dx}+a_0(x)y_2=0$$

Adding ${\alpha }_1$ times the first equation to ${\alpha }_2$ times the second equation gives

$$a_k(x)\frac{d^k({\alpha }_1{y}_1+{\alpha }_2{y}_2)}{d{x}^k}+a_{k-1}(x)\frac{d^{k-1}({\alpha }_1{y}_1+{\alpha }_2{y}_2)}{d{x}^{k-1}}+\cdots +a_1(x)\frac{d({\alpha }_1{y}_1+{\alpha }_2{y}_2)}{dx}+a_0(x)({\alpha }_1{y}_1+{\alpha }_2{y}_2)=0$$

Hence ${\alpha }_1{y}_1+{\alpha }_2{y}_2$ is also a solution to the ODE

Note that this holds as differentiation is a Linear Map

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02 - General Solution to a Inhomogeneous Linear ODE

General Solution to a Inhomogeneous Linear ODE

Let $y_{p}x)$ be a solution, aka the particular integral, of the inhomogeneous ODE

$$a_k(x)\frac{d^{k}y}{d{x}^k}+a_{k-1}(x)\frac{d^{k-1}y}{d{x}^{k-1}}+\cdots +a_1(x)\frac{dy}{dx}+a_0(x)y=f(x)$$

Such that $y=y_p$ satisfies the equation above

Then a function $y(x)$ is a solution fo the inhomogeneous linear ODE
If and only if $y(x)$ is in the form of

$$y(x)=y_c(x)+y_p(x)$$

Where $y_c(x)$ is a solution to the corresponding homogenous linear ODE, that is

$$a_k(x)\frac{d^{k}y}{d{x}^k}+a_{k-1}(x)\frac{d^{k-1}y}{d{x}^{k-1}}+\cdots +a_1(x)\frac{dy}{dx}+a_0(x)y=0$$

With $y_c(x)$ being known as the complementary function

Proof

If $y(x)=y_c(x)+y_p(x)$ is a solution to the ODE then

$$a_k(x)\frac{d^k(y_c+y_p)}{d{x}^k}+a_{k-1}(x)\frac{d^{k-1}(y_c+y_p)}{d{x}^{k-1}}+\cdots +a_1(x)\frac{d(y_c+y_p)}{dx}+a_0(x)(y_c+y_p)=f(x)$$

Hence

$$\left(a_k(x)\frac{d^k{y}_c}{d{x}^k}+\cdots +a_0(x)y_c\right)+\left(a_k(x)\frac{d^k{y}_p}{d{x}^k}+\cdots +a_0(x)y_p\right)=f(x)$$

As the second bracket equals $f(x)$ as $y_p(x)$ is a solution to the ODE then
The first bracket must equal $0$ hence

$$y_c(x)\text{ is a solution of the homogeneous ODE}$$

Note that the particular integral is found via guess work (generally similar to $f(x))$

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3.2 Second Order Homogeneous Linear ODEs

Solving Second Order Homogeneous Linear Differential Equations

Suppose $z(x)\ne 0$ then

$$p(x)\frac{d^{2}y}{d{x}^2}+q(x)\frac{dy}{dx}+r(x)y=0$$

Using substitution

$$y(x)=u(x)z(x)$$

such that

$$\frac{dy}{dx}=\frac{du}{dx}z+u\frac{dz}{dx}\text{ and }\frac{d^{2}y}{d{x}^2}=\frac{d^{2}u}{d{x}^2}+2\frac{du}{dx}\frac{dz}{dx}+u\frac{d^{2}z}{d{x}^2}$$

Then substituting these values into the ODE then

$$p(x)(u^{\prime \prime }z+2u^{\prime }{z}^{\prime }+u{z}^{\prime \prime })+q(x)(u^{\prime }z+u{z}^{\prime })+r(x)uz=0$$

As $z$ is a solution to the ODE then $p(x)z^{\prime \prime }+q(x)z^{\prime }+r(x)z=0$ hence

$$p(x)z{u}^{\prime \prime }+(2p(x)z^{\prime }+q(x)z)u^{\prime }=0$$

This is now a homogenous differential equation of first order for $u^{\prime }$ which should be solvable hence there is a general solution

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3.3 Linear ODEs with constant coefficients

3.3.1 The Homogenous Case

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

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3.3.2 The Inhomogeneous Case

Finding the Solution to the Inhomogeneous Linear Differential Equation

The particular solution $y_p(x)$ is generally found by trial and error by mimicking the form of $f(x)$

Particular Solution similar to Complementary Function $x$

Multiply the previous particular solution by

Or when you solve the constant coefficients which then make $y_p(x)=0$

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Example 1 - Particular Solution

Find the particular solution of

$$\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+2y=x$$

Solution

Try $y_p(x)=ax+b$ where $a,b\in \mathbb{R}$

So we get

$$\frac{d{y}_p}{dx}=a\text{ and }\frac{d^2{y}_p}{d{x}^2}=0$$

Substituting back we get

$$0-3a+2(ax+b)=x$$

As this holds for all $x$ then comparing coefficients

$$\begin{array}{rl}2a& =1⟹a=\frac{1}{2}\\ -3a+2b& =0⟹b=\frac{3}{4}\end{array}$$

Hence

$$y_p(x)=\frac{x}{2}+\frac{3}{4}$$

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Example 2 - Particular Solution

Find the particular solution of

$$\frac{d^{2}y}{d{x}^2}+4\frac{dy}{dx}+4y=\sin (3x)$$

Solution

The auxiliary equation is $m^2+4m+4$ which has repeated root $-2$
Hence the complementary function is

$$y_c(x)=(C_{1}x+C_2)e^{-2x}$$

For the particular solution we will pick something like

$$y_p(x)=a\cos (3x)+b\sin (3x)$$

Differentiating and substituting gives you the equations

$$\begin{array}{rl}-9a+12b+4a& =0\\ -9b-12a+4b& =1\end{array}$$

Hence

$$a=-\frac{12}{169}\text{ and }b=-\frac{5}{169}$$

So we get particular solution

$$y_p(x)=-\frac{12}{169}\cos (3x)-\frac{5}{169}\sin (3x)$$

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Example 3 - Particular Solution (Complementary Function Clash)

Fid the particular solution of

$$\frac{d^{2}y}{d{x}^2}-6\frac{dy}{dx}+9y=e^{3x},y(0)=y^{\prime }(0)=1$$

Solution $m^2-6y+9=0$ leads to the complementary function

The auxiliary equation

$$y_c(x)=(C_{1}x+C_2)e^{3x}$$

Our first guess for the particular solution is $a{e}^{3x}$
However as this is contained in $y_c(x)$ then we try $ax{e}^{3x}$
However again this again is also contained in $y_c(x)$

Hence our particular solution is in the form $a{x}^2{e}^{3x}$
Differentiating and substituting into our ODE gives that $\alpha =\frac{1}{2}$

Hence the particular solution is

$$y_p(x)=\frac{1}{2}{x}^2{e}^{3x}$$

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