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Mathematics is the language in which God has written the universe, Galileo Galilei
Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y' + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$
It involves (e.g., $\frac{dy}{dx} = 3x +5y$):
In mathematics, a linear ordinary differential equation (ODE) is an equation involving an unknown function and its derivatives, characterized by the property that both the function and its derivatives appear linearly. This means that the unknown function and its derivatives are raised only to the first power. They are never multiplied together, nor are they composed within nonlinear functions (such as $\sin(y)$ or $e^y$). The general form for a first-order linear ODE is: a(x)y’ + b(x)y = c(x).
In mathematics, a second-order linear homogeneous differential equation ODE with constant coefficients is an equation of the form: ay'' + by' + cy = 0 where:
This equation is characterized by:
This type of differential equation arises frequently in physics and engineering, modeling systems such as mechanical vibrations (mass-spring-damper) and electrical circuits (LRC).
When solving ordinary differential equations (ODEs), our primary goal is to find the general solution, which represents the complete family of all possible valid solutions that satisfy the equation.
For a second-order linear homogeneous ODE, the general solution is defined as: $\boxed{y(x) = c_1y_1(x) + c_2y_2(x)}$ where:
Two solutions, $y_1$ and $y_2$, are considered linearly independent if neither is a constant multiple of the other. Mathematically, the only way the equation $c_1y_1(x) + c_2y_2(x) = 0$ can be true for all $x$ is if both $c_1 = 0$ and $c_2 = 0$.
Why is this property so crucial?
Completeness: Only linearly independent solutions guarantee that the general solution encompasses all possible solutions to the ODE. If the solutions are linearly dependent (e.g., $y_2 = 3y_1$), they collapse into a single mathematical dimension and do not span the entire solution space.
Uniqueness: Linearly independent solutions allow us to satisfy arbitrary initial conditions independently. If the solutions are dependent, we simply do not have enough mathematical freedom to adjust the constants $c_1$ and $c_2$ to fit specific real-world initial states.
One common method to test for linear independence is using the Given two differentiable functions $y_1(x)$ and $y_2(x)$, the Wronskian determinant is defined as: $W(y_1, y_2)(x) = \begin{vmatrix} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \end{vmatrix} = y_1(x)y_2'(x) - y_1'(x)y_2(x)$
Linear Independence Criterion: If the Wronskian $W(y_1, y_2)(x)$ is NOT identically zero on the interval of interest (i.e., W ≠ 0 for at least some x in the interval), then $y_1$ and $y_2$ are linearly independent. on that interval.
The Superposition Principle is the fundamental concept that allows us to construct general solutions for linear homogeneous differential equations.
It states: If $y_1(x)$ and $y_2(x)$ are solutions to a linear homogeneous ODE, then any linear combination of these solutions, $y(x) = c_1y_1(x) + c_2y_2(x)$, is also a valid solution for any constants $c_1$ and $c_2$.
This principle underpins the structure of the solution space for linear homogeneous ODEs. Because of this, the set of all possible solutions forms a vector space with a dimension equal to the order of the differential equation (which is exactly two in our case).
Below are two distinct ways to formally prove this principle.
Consider the general second-order linear homogeneous ODE: $y'' + p(x)y' + q(x)y = 0$
Suppose $y_1$ and $y_2$ are known, valid solutions. This means they independently satisfy the ODE:
$y_1'' + p(x)y_1' + q(x)y_1 = 0$
$y_2'' + p(x)y_2' + q(x)y_2 = 0$
Let’s test a linear combination $y(x) = c_1y_1 + c_2y_2$ to see if it also satisfies the equation.
We can also prove this by reframing the ODE using the Differential Operator, $D$. Let’s rewrite $y'' + p(x)y' + q(x)y = 0$ as: $(D^2 + p(x)D + q(x))y = 0$.
We can bundle this entire expression into a single linear operator called L: $L = D^2 + p(x)D + q(x)$.
So our differential equation simplifies elegantly to: $L(y) = 0$
Think of $L$ as a “black box” function: you input a function $u(x)$, and it outputs a new function $v(x)$. Because $L$ is a linear operator, it inherently satisfies two strict mathematical properties:
Note: The standard differential operator naturally possesses these properties: $(u_1 + u_2)' = u_1' + u_2'$ and $(cu)' = cu'$. Since $L$ is just made of standard derivatives, it inherits these rules.
Additivity. Suppose $u_1$ and $u_2$ are two functions. Then, the operator L applied to the sum of $u_1$ and $u_2$ is:
L($u_1$ + $u_2$) = ($u_1$ + $u_2$)’’ + p(x)($u_1$ + $u_2$)’ + q(x)($u_1$ + $u_2$)
Using the basic properties of derivatives and addition, we can split each term:
L($u_1$ + $u_2$) = $u_1$’’ + $u_2$’’ +p(x)($u_1$’ + $u_2$’) + q(x)($u_1$ + $u_2$)
Now, rearrange terms:
L($u_1$ + $u_2$) = ($u_1$’’ + p(x)$u_1$’ + q(x)$u_1$) + ($u_2$’’ + p(x)$u_2$’ + q(x)$u_2$) = L($u_1$) + L($u_2$).
Homogeneity (Scaling). Next, we need to show that L satisfies the scaling property. Suppose c is a constant and u is a function. Then applying L to the constant multiple c⋅u gives: L(c·u) = (c⋅u)′′ +p(x)(c⋅u)′ +q(x)(c⋅u)
Since the derivative of a constant multiple of a function is just the constant times the derivative of the function, we can factor out the constant c from each term:
L(c⋅u) = c⋅u′′ +p(x)⋅c⋅u′ +q(x)⋅c⋅u =[Factor c out] = c·(u’’ + p(x)u’ + q(x)) = c·L(u).
Our ODE is simply $L(y) = 0$. If $y_1$ and $y_2$ are solutions, then by definition $L(y_1) = 0$ and $L(y_2) = 0$.
Let’s plug our linear combination $c_1y_1 + c_2y_2$ into the operator: $L(c_1y_1 + c_2y_2)$
By the Additivity property, we can split this: $L(c_1y_1 + c_2y_2) = L(c_1y_1) + L(c_2y_2)$
By the Homogeneity property, we can pull the constants out: $L(c_1y_1) + L(c_2y_2)$ = c_1L(y_1) + c_2L(y_2)$
Since we already know $L(y_1) = 0$ and $L(y_2) = 0$, we substitute those in: $= c_1(0) + c_2(0) = 0$
Because $L(c_1y_1 + c_2y_2) = 0$, the linear combination is a valid solution. $\blacksquare$
How do we know there isn’t a secret third solution out there?
For second-order linear homogeneous ODEs:
To fully grasp how to solve second-order linear homogeneous ODEs, it is helpful to compare an equation with constant coefficients to one with variable coefficients. The strategies for finding the initial roots differ, but the overarching principles (finding linearly independent solutions and verifying them via the Wronskian) remain the same.
Find the characteristic Equation. For linear homogeneous ODEs with constant coefficients, we can assume a solution of the form: $y(x) = e^{rx}$ where r is a constant to be determined. Substitute $y(x) = e^{rx}$, $y'(x) = re^{rx}$, and $y''(x) = r^2e^{rx}$ into the ODE: $(r^2e^{rx})-5(re^{rx})+ 6(e^{rx}) = 0$.
Since the exponential term $e^{rx}$ is never zero, we can divide the entire equation by it to obtain the characteristic equation: $r^2 - 5r + 6 = 0$
Solve the characteristic equation. Factor the quadratic equation to find the roots: $(r - 2)(r - 3) = 0$. The roots are distinct and real: $r = 2$ and $r = 3$.
Write the General Solution. Second-order linear ODEs: Real and distinct roots. Because we have two distinct real roots, the solution is a linear combination of two exponential terms. The general solution is: $y(x) = c_1e^{2x} + c_2e^{3x}$.
Here, $y_1(x) = e^{2x}$ and $y_2(x) = e^{3x}$ are our foundational solutions, and $c_1$ and $c_2$ are arbitrary constants determined by initial conditions.
Verify Linear Independence. Compute the Wronskian to mathematically confirm that these two solutions are linearly independent:: $W(y_1, y_2)(x) = |\begin{smallmatrix}y_1(x) & y_2(x)\\\ y_1'(x) & y_2'(x)\end{smallmatrix}| = y_1(x)y_2'(x) -y_1'(x)y_2(x) = e^{2x}\cdot 3e^{3x}-2e^{2x}\cdot e^{3x} = 3e^{5x}-2e^{5x}=e^{5x}≠0$. Since the Wronskian is non-zero for all x, the solutions are linearly independent.
This specific form, where the power of $x$ matches the order of the derivative in each term, is known as a Cauchy-Euler equation.
Step 1. Assume a Solution Form.
For Euler equations, standard exponentials do not work. Instead, we assume a polynomial solution of the form $y(x) = x^r$ where r is a constant to be determined.
Step 2. Compute the first and second derivatives and substitute into the ODE.
Find the first and second derivatives: $y' = rx^{r-1}$, $y'' = r(r-1)x^{r-2}$
Substitute these back into the original ODE: $x^2\big(r(r-1)x^{r-2}\big) - x\big(rx^{r-1}\big) + x^r = 0$
$r(r-1)x^r - rx^r + x^r = 0$
Step 3. Simplify to the Characteristic Equation: Factor out the common $x^r$ term: $[r(r-1) - r + 1]x^r = 0$
Since we are evaluating for $x > 0$, $x^r \neq 0$. We can divide both sides by $x^r$ to yield the characteristic equation for this Cauchy-Euler ODE: $r^2 - 2r + 1 = 0$
Step 4. Solve the characteristic Equation. Factoring the quadratic reveals a repeated real root: $(r - 1)^2 = 0 \implies r = 1$.
Our first valid solution is therefore $y_1(x) = x^1 = x$.
Step 5. Find the General Solution. (Reduction of Order).
Normally, when the characteristic equation has distinct roots $r_1, r_2$ the general solution is $y(x) = c_1x^{r_1} + c_2x^{r_2}$.
For constant coefficient differential equations, when we have repeated roots $r_0$ the general solution is $y(t) = c_1e^{r_0t} + c_2te^{r_0t}$
Here, for a variable coefficient equation, we must use the Reduction of Order method to find the second solution.
We assume the second solution takes the form $y_2(x) = y_1(x)v(x) = xv(x)$. We must find $v(x)$ such that $y_2(x)$ satisfies the ODE.
Compute the derivatives using the product rule: $y_2' = v'x + v$, $y_2'' = v''x + v' + v' = v''x + 2v'$.
Substitute $y_2$, $y_2'$, and $y_2''$ into the original ODE ($x^2y'' - xy' + y = 0$) and we get: $x^2(v''x + 2v') - x(v'x + v) + (xv) = 0$
Expand and combine like terms: $x^3v'' + 2x^2v' - x^2v' - xv + xv = 0$. It beautifully simplifies to: $x^3v'' + x^2v' = 0$.
Divide the entire equation by $x^2$ (which is valid since $x > 0$): $xv'' + v' = 0$
This is a first-order linear ODE with respect to v′(x). To solve it, we make a substitution: let $u = v'$. $xu' + u = 0 \implies x\frac{du}{dx} = -u \implies \frac{du}{dx} = -\frac{u}{x}$
This is a separable equation. Separate the variables and integrate both sides: $\frac{du}{u} = \frac{-1}{x}dx$. $\frac{du}{u} = -\frac{dx}{x} \implies \ln|u| = -\ln|x| + C$
$u = e^{-\ln|x| + C} = e^{\ln(x^{-1})} \cdot e^C = \frac{K}{x}$ where $K = e^C$ is a constant.
Since $u = v'$, we have $v'(x) = \frac{K}{x}$. Integrating again gives: $v(x) = K\ln(x) + C'$.
Because we are just looking for a single independent baseline solution to pair with our arbitrary constants later, we can choose the simplest non-zero version of this function by setting $K = 1$ and $C' = 0$. Therefore, $v(x) = \ln(x)$.
Therefore, our second solution is $y_2(x) = xv(x) = x\ln(x)$.
Step 6. Verify Linear Independence.
To ensure $y_1(x) = x$ and $y_2(x) = x\ln(x)$ form a fundamental set of solutions, calculate the Wronskian:
$y_1 = x \implies y_1' = 1$, $y_2 = x\ln(x) \implies y_2' = x\left(\frac{1}{x}\right) + \ln(x) = 1 + \ln(x)$
$W(y_1, y_2)(x) = (x)(1 + \ln(x)) - (1)(x\ln(x)) = x + x\ln(x) - x\ln(x) = x$.
Since we restricted the domain to $x > 0$, the Wronskian $W = x \neq 0$. The solutions are linearly independent.
Step 7. Write the General Solution. General Solution: $y(x) = c_1x^r + c_2x^r\ln(x)$.
Substituting r=1, we get: $y(x) = c_1x + c_2x\ln(x)$ where x and xlnx are linearly independent solutions and $c_1$ and $c_2$ are constants to be determined by the initial or boundary conditions.
Step 8. Apply Initial Conditions. If given the initial conditions $y(x_0) = a$ and $y'(x_0) = b$, we can solve for $c_1$ and $c_2$.
Using the general solution and its derivative $y'(x) = c_1 + c_2(\ln(x) + 1)$, we construct a system of linear equations:
$\begin{cases} c_1x_0 + c_2x_0\ln(x_0) = a \\ c_1 + c_2(\ln(x_0)+1) = b \end{cases}$
Because the Wronskian evaluated at $x_0$ is non-zero ($W(x_0) = x_0 \neq 0$), linear algebra guarantees that this system will always yield a unique, solvable pair of values for $c_1$ and $c_2$
Recall. A $2\times 2$ linear system has a unique solution if and only if the coefficient matrix is invertible, i.e., its determinant is nonzero.
Compute the determinant: $\det = x_0(\ln(x_0)+1) - x_0\ln(x_0) = x_0.$ But this determinant is exactly the Wronskian evaluated at $(x_0)$.
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