03 - Homogenous ODEs

Basis of Solutions to $(H)$

Let $(H)$ denote the Homogenous Equation ($Ly=0$)

Let $y_1$ and $y_2$ be two particular solutions of $(H)$ satisfying initial conditions at some $x=a$

$$y_1(a)=1,y_1^{\prime }(a)=0,y_2(a)=0,y_2^{\prime }(a)=1$$

Then if $y(x)$ is a solution of $(H)$ then

$$y(x)\text{ is a linear combination of }{y}_1\text{ and }{y}_2$$

Proof

By Picard’s Theorem both $y_1(x)$ and $y_2(x)$ exist
Both are unique at least in a neighbourhood of $x=a$ provided $P_2(a)\ne 0$

Wronskian has $W=1$ at $x=a$ so is non-zero in the same neighbourhood of $x=a$
Hence they are linearly independent

Suppose $y(x)$ is any other solution of $(H)$ so set

$$Y(x)=y_1(x)y(a)+y_2(x)y^{\prime }(a)$$

So $Y(x)$ is a solution of $(H)$ and satisfies initial conditions

$$Y(a)=y(a),Y^{\prime }(a)=y^{\prime }(a)$$

By uniqueness then $Y(x)\equiv y(x)$ so $y(x)$ is a linear combination of $y_1$ and $y_2$
Hence $y_1$ and $y_2$ spans the vector space of solution $Y$, hence a basis

Properties of Space of Solutions of $(H)$

1. Dimension of the space of solutions of $(H)$ is $2$

2. Any pair of solutions of $(H)$ with $W\ne 0$ is a basis

Example

Let $(H)$ denote the Homogenous Equation ($Ly=0$)
If $P_2,P_1,P_0$ are constant then $(H)$ has exponential solutions in form of

$$y=e^{mx}$$

where $m$ satisfies quadratic equation (also known as the auxiliary equation)

$$P_2{m}^2+P_{1}m+P_0=0$$

If roots are repeated or complex then care must be taken

Example

Coefficients are in the form of

$$P_2(x)=\alpha x^2,P_1(x)=\beta x,P_0(x)=\gamma$$

where $\alpha ,\beta ,\gamma$ are constants

So $(H)$ is in form

$$\alpha x^2\frac{d^{2}y}{d{x}^2}+\beta x\frac{dy}{dx}+\gamma y=0$$

Note that the powers of each for each term is the same

Hence solutions are in the form

$$y(x)=x^m$$

where $m$ satisfies quadratic equation

$$\alpha m(m-1)+\beta m+\gamma =0$$

If roots are repeated or complex then care must be taken

Alternate Substitution (Equivalent)

Substitute $x=e^t$ to get constant-coefficients equation

$$\alpha \frac{d^{2}y}{d{t}^2}+(\beta -\alpha )\frac{dy}{dt}+\gamma y=0$$

Example

If one solution $y_1(x)$ is known then
General Solution can be found by solving an ODE of reduced order

Express solution to ODE $(H)$ in form of

$$y(x)=v(x)y_1(x)$$

Substituting into $(H)$ then using the fact that $y_1$ is a solution of $(H)$ obtain

$$P_2{y}_1{v}^{\prime \prime }+(2P_2{y}_1^{\prime }+P_1{y}_1)v^{\prime }=0$$

which is a separable first-order ODE for $v^{\prime }$ with solution

$$v^{\prime }(x)=\frac{\text{const}}{y_1(x{)}^2}\exp \left(-\int \frac{P_1(x)}{P_2(x)}dx\right)$$

Integrating once more gives $v(x)$ and hence general solution $y(x)=v(x)y_1(x)$

Alternative Derivation from Wronskian

Constructing the general solution from a single known solution can also be derived from Wronskian

$$\begin{array}{rl}W(x)& =y_1(x)y_2^{\prime }(x)-y_2(x)y_1^{\prime }(x)\\ & =y_1(x{)}^2\frac{d}{dx}\left(\frac{y_2(x)}{y_1(x)}\right)\\ & =\text{ const }\cdot \exp \left(-\int \frac{P_1(x)}{P_2(x)}dx\right)\end{array}$$

Hence we can construct $y_2(x)$ given $y_1(x)$

$n$th Order Homogenous Linear ODEs

ODEs in form

$$Ly(x)=y^{(n)}(x)+P_{n-1}(x)y^{(n-1)}(x)+\cdots +P_1(x)y^{\prime }+P_0$$

(x)y(x) = 0

<small>Note that the highest-order derivative as coefficient $1$ (divided through as homogenous)</small>