A second order linear ODE is of the form \begin{align} \difftwo{y} + p(x)\diffone{y} + q(x)y = r(x) \end{align}
Any second order ODE not expressible in this form will be non-linear.
For any homogenous second order linear ODE, any linear combination of two solutions is also a solution. In particular, this means that sums, and sums after multiplying with constants are also solutions of the ODE. This result is only valid for linear homogenous equations and not non-homogenous ones.
This can be verified by first assuming two solutions $y_{1}$ and $y_{2}$, and then substituiting a linear combination $c_{1}y_{1} + c_{2}y_{2}$ in the same homogenous equation. This linear combination is called the general solution to the ODE.
Hence, while evaluating $y_{1}$ and $y_{2}$, we will not keep any constants, since they will automatically appear when preparing the general solution.
The two solutions $y_{1}$ and $y_{2}$ form the basis or a fundamental system of solutions of the ODE, and must be linearly independent. That is, the two functions must satisfy \begin{align} k_{1}y_{1} + k_{2}y_{2} = 0 \forall x \in I \implies k_{1} = 0 \;\text{and}\; k_{2} = 0 \end{align} where $I$ is the solution interval under consideration.
If the solutions were linearly dependent, there would be a non-trivial solution for the above equation, implying that the two solutions are linearly dependent. The two functions $y_{1}$ and $y_{2}$ are called the basis as the general solution is the set of all functions that are linear combinations of the two bases.
The linear independence of two solutions can also be verified by checking that their ratio is not a constant (but a function of the independent variable).
Similar to linear ODEs, second order linear homogenous equations have an IVP that can be stated with two initial values, one on the function $y$ itself, and the other on its derivative: $y(x_{0}) = K_{0}$ and $\diffone{y}(x_{0}) = K_{1}$. The two conditions can be used to determine the two constants in a genral solution: $y = c_{1}y_{1} + c_{2}y_{2}$.
The unique solution will pass through the point $(x_{0}, K_{0})$ with the tangent $K_{1}$, and is called a particular solution.
This method for second order linear homogenous equations relies on already knowing one solution of the equation. Then, we determine the second using the fact that the two solutions will be linearly independent.
Suppose $y_{1}$ is one solution, then $y_{2} = uy_{1}$ where $u$ is a function of $x$ (and not a constant), is the second solution. Substituiting in our ODE, \begin{align} \difftwo{y} + p(x)\diffone{y} + q(x)y &= 0\newline \difftwo{y_{1}} + p(x)\diffone{y_{1}} + q(x)y_{1} &= 0\newline \difftwo{y_{2}} + p(x)\diffone{y_{2}} + q(x)y_{2} &= 0\newline \implies \difftwo{(uy_{1})} + p(x)\diffone{(uy_{1})} + q(x)(uy_{1}) &= 0\newline \diffone{(uy_{1})} &= \diffone{u}y_{1} + u\diffone{y_{1}}\newline \difftwo{(uy_{1})} &= \difftwo{u}y_{1} + 2\diffone{u}\diffone{y_{1}} + u\difftwo{y_{1}}\newline \implies \difftwo{u}y_{1} + 2\diffone{u}\diffone{y_{1}} + u\difftwo{y_{1}} + p(x)(\diffone{u}y_{1} + u\diffone{y_{1}}) + q(x)uy_{1} &= 0\newline \text{Rearranging,}: \difftwo{u}y_{1} + \diffone{u} \roundbr{2\diffone{y_{1}} + p(x)y_{1}} + u \roundbr{\difftwo{y_{1}} + p(x)\diffone{y_{1}} + q(x)y_{1}} = 0\newline \implies \difftwo{u}y_{1} + \diffone{u} \roundbr{2\diffone{y_{1}} + p(x)y_{1}} = 0\newline \end{align}
The last equation follows since $y_{1}$ is a solution. Rearranging, \begin{align} \frac{\difftwo{u}}{\diffone{u}} &= -\frac{2\diffone{y_{1}} + p(x)y_{1}}{y_{1}} = -2\frac{\diffone{y_{1}}}{y_{1}} + p(x)\newline \implies \int \frac{d\roundbr{\diffone{u}}}{\diffone{u}} &= -2\int \frac{d\roundbr{y_{1}}}{y_{1}} - \int p(x) dx\newline \ln \diffone{u} &= -2 \ln y_{1} - \int p(x) dx\newline \diffone{u} &= \frac{e^{-\int p(x) dx}}{y_{1}^{2}}\newline u &= \int \frac{e^{-\int p(x) dx}}{y_{1}^{2}} dx \end{align}
which must not be a constant since $y_{1}$ and $y_{2}$ are linearly independent. Now, $y_{2}$ is simply $uy_{1}$.
We consider equations of the form $\difftwo{y} + a\diffone{y} + by = 0$.
We know that the first order linear equation $\diffone{y} + ay = 0$ has a solution of the form $e^{-ax}$. We can try to substitute a similar solution $e^{\lambda x}$ in the equation above to get \begin{align} e^{\lambda x}\roundbr{\lambda^{2} + a\lambda + b} = 0 \end{align}
Since the exponent is non-zero, the second part of the above equation must be zero. This equation is called the characteristic equation. This is a quadratic in $\lambda$ and the roots are \begin{align} \lambda = -\frac{-a \pm \sqrt{a^{2} - 4b}}{2} \end{align} has the following three possibilities for the roots:
The two roots are complex Recall Euler’s theorem \begin{align} e^{ix} = \cos x + i\sin x \end{align} The two complex roots of the quadratic equation will be \begin{align} \lambda = -\frac{-a \pm i\sqrt{4b - a^{2}}}{2} \end{align} Let $\sqrt{4b - a^{2}}/2 = \omega$. Similar to the real and distinct roots case, we can write the two solutions as \begin{align} y_{1} &= e^{\lambda_{1}x} = e^{-ax/2}\roundbr{\cos \omega x + i\sin \omega x}\newline y_{2} &= e^{\lambda_{2}x} = e^{-ax/2}\roundbr{\cos \omega x - i\sin \omega x} \end{align}
We can take a linear combination of the above two roots to get a new basis set (whose ratio will not be a constant) by
Instead of using $dy/dx$, we can also write this as $Dy$ where $D$ is the differential operator. Higher order derivatives like $d^{2}y/dx^{2}$ then become $D^{2}y$ and so on.
This operator can be treated like any other algebraic operator. Let $I$ denote the identity operator such that $Iy = y$. Also, let $L$ denote our linear differential equation. $L$ has the property
$P(D)$ denotes our operator for this equation, and can be treated like a operator polynomial. This way, the operators can be treated like any other polynomial similar to algebra.
We can factorize this polynimal to find two distinct roots, exactly like solving the distinct roots case. This approach of working is more common in engineering settings.
Consider Linear ODEs of the form \begin{align} x^{2}\difftwo{y} + ax\diffone{y} + by = 0 \end{align} where $a$ and $b$ are constants. Assuming $y = x^{m}$ as the solution,
\begin{align} x^{2}m(m-1)x^{m-2} + axmx^{m-1} + bx^{m} &= 0\newline \implies m(m-1) + am + b &= 0\newline m^{2} + (a - 1)m + b &= 0 \end{align}
Therefore, $m$ must be a root of the above quadratic equation. There are three possible cases
Complex Roots The roots can be denoted by $(1-a)/2 + i\omega$ where $\omega = (4b - (a-1)^{2})/2$. Taking cue from the real distinct roots scenario, the solutions will be \begin{align} y_{1} &= x^{(1-a)/2 + i\omega} = x^{(1-a)/2} (e^{\ln x})^{i\omega} = x^{(1-a)/2} e^{i\omega \ln x}\newline &= x^{(1-a)/2}\roundbr{\cos \roundbr{\omega\ln x} + i\sin\roundbr{\omega\ln x}}\newline y_{2} &= x^{(1-a)/2}\roundbr{\cos \roundbr{\omega\ln x} - i\sin\roundbr{\omega\ln x}}\newline \end{align}
Similar to how a real basis was derived for Second Order Linear ODE with Constant Coefficients in the complex roots of characeristic equation case, the two solutions will be \begin{align} y_{1} &= x^{(1-a)/2}\cos \roundbr{\omega\ln x}\newline y_{2} &= x^{(1-a)/2}\sin \roundbr{\omega\ln x}\newline y &= x^{(1-a)/2}\roundbr{c_{1}\cos \roundbr{\omega\ln x} + c_{2}\sin\roundbr{\omega\ln x}} \end{align}
Typically, the complex roots case is of no real practical value.
The Wronskian of the solutions is defined as \begin{align} W = y_{1}\diffone{y_{2}} - y_{2}\diffone{y_{1}} = \detm{\begin{matrix}y_{1} &y_{2}\newline \diffone{y_{1}} &\diffone{y_{2}} \end{matrix}} \end{align}
The Wronskian plays an important role in determining the independence of solutions for a homogenous second order linear differential equation. For the open interval $I$ under consideration, the solutions are linearly dependent if there exists $x_{0} \in I$ such that $W(x_{0}) = 0$. Further, if the Wronskian is 0 for an $x_{0} \in I$, then it is zero throughout. Thus, $W(x) \neq 0$ for any $x_{1} \in I$, then the solutions are linearly independent and form a basis on $I$.
A second order linear homogenous ODE has no singular solutions. That is, all solutions can be obtained from the general solution.
We consider equations of the form \begin{align} \difftwo{y} + p(x)\diffone{y} + q(x)y = r(x) \end{align}
where $r(x) \neq 0$. The general solution to this ODE is defined as a combination of the solution to the corresponding homogenous solution ($y_{h}$), and a solution to the complete nonhomogenous equation ($y_{p}$, containing no arbitrary constants). \begin{align} y_{h} &= c_{1}y_{1} + c_{2}y_{2}\newline y &= y_{h} + y_{p} \end{align}
A particular solution is obtained by evaluating the constants $c_{1}$ and $c_{2}$ based on the initial conditions.
This method applies to nonhomogenous linear ODEs with constant coefficients \begin{align} \difftwo{y} + a\diffone{y} + by = r(x) \end{align}
When $r(x)$ is a common function like sinusoidal, exponential, polynomial (these all are functions whose derivatives are of similar form to the function itself), or a combinatation of these, the derivatives will also be of similar form. Hence, we can choose the solution to be a combination of such functions with undetermined coefficients and put the solution in the equation to determine those coefficients.
We can follow the following rules to determine the solution
Basic Rule: If $r(x)$ is in the left column of the following table, a suitable $y_{p}(x)$ can be picked from the right column.
| $r(x)$ | $y_{p}(x)$ |
|---|---|
| $ke^{\gamma x}$ | $Ce^{\gamma x}$ |
| $kx^{n}$ $n=0,1,\ldots$ | $K_{n}x^{n} + K_{n-1}x^{n-1} + \cdots + K_{1}x + K_{0}$ |
| $k\sin \omega x$ or $k\cos \omega x$ | $K\sin \omega x + M\cos \omega x$ |
| $ke^{\alpha x}\sin \omega x$ or $ke^{\alpha x}\cos \omega x$ | $e^{\alpha x}\roundbr{K\sin \omega x + M\cos \omega x}$ |
In the above procedure, we will first look for solution to the homogenous equation. Furthermore, an incorrect choice for the coefficients will lead to a contradiction. The constants assumed in $y_{p}$ can be found by comparing the left and right side of the ODE term by term after substituiting the solution (for instance, the coefficient of $x^{2}$ should be the same on both the sides).
An important points from an engineering perspective needs to be considered here. If the roots of the homogneous equation are negative, or have negative real parts, only then will the nonhomogenous equation will have a stable solution. This implies $y_{h} \to 0$ as $x \to \infty$. In this case the general solution will approach a transient state and the steady state value of $y_{p}(x)$.
This method is credited to Lagrange and applies to any function as coefficients that is continuous. Assuming $y_{1}$ and $y_{2}$ are the solutions to the homogenous equation, $y_{p}$ is given by \begin{align} \difftwo{y} + p(x)\diffone{y} + q(x)y &= r(x)\newline y_{p} &= -y_{1}\int\frac{y_{2}r}{W}dx + y_{2}\int\frac{y_{1}r}{W}dx\newline W &= y_{1}\diffone{y_{2}} - y_{2}\diffone{y_{1}} \end{align}
where $W$ is called the Wronskian. Remember that this formulation is valid for the ODE in its standard form where the coefficient of $\difftwo{y}$ is 1. If that is not the case, we can divide the equation throughout by the coefficient to make it unity.
This method can be derived by assuming $y_{p}$ to be a combination of the solutions of the homogenous equation, but using functions of $x$ instead of constants. Then, after substituiting the solution in the ODE, we try to derive the values of these functions (coefficients).