In calculus, we study how to differentiate a given function $f$ to find its derivative $f'(x)$, as well as how to find a function whose derivative is a given function $g(x)$. This inverse process leads to equations of the form $\frac{dy}{dx} = g(x)$, which are the simplest examples of differential equations.
The classification of differential equations by order and degree is fundamental to determining their solution methodologies.
| Differential Equation | Highest Order Derivative | Polynomial in Derivatives? | Order | Degree |
|---|---|---|---|---|
| $$\frac{dy}{dx} = e^x$$ | $$\frac{dy}{dx}$$ | Yes | 1 | 1 |
| $$\frac{d^2y}{dx^2} + y = 0$$ | $$\frac{d^2y}{dx^2}$$ | Yes | 2 | 1 |
| $$\left(\frac{d^3y}{dx^3}\right) + x^2\left(\frac{d^2y}{dx^2}\right)^3 = 0$$ | $$\frac{d^3y}{dx^3}$$ | Yes | 3 | 1 |
| $$\left(\frac{dy}{dx}\right)^2 + \left(\frac{dy}{dx}\right) - \sin^2 y = 0$$ | $$\frac{dy}{dx}$$ | Yes | 1 | 2 |
| $$\frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0$$ | $$\frac{dy}{dx}$$ | No (due to $\sin(y')$) | 1 | Not Defined |
Mathematical Problem Statement: Find the order and degree (if defined) of the differential equation: $$y''' + y^2 + e^{y'} = 0$$
Step-by-Step Algorithmic Solution Trace:
- Identify the highest order derivative present in the equation. The term $y'''$ represents the third derivative $\frac{d^3y}{dx^3}$, which is the highest order derivative. Thus, the order is 3.
- Examine if the differential equation is a polynomial equation in its derivatives. The term $e^{y'}$ is an exponential function of a derivative, which means the equation cannot be written as a polynomial in terms of $y', y'', y'''$.
- Since it is not a polynomial equation in its derivatives, the degree is not defined.
Final Evaluation Output: Order = 3, Degree = Not Defined
A solution of a differential equation is a relation between the variables (not involving derivatives) which satisfies the given differential equation.
Historical Chronicle: Henri Poincaré (1854-1912)
Mathematical Problem Statement: Verify that the function $y = e^{-3x}$ is a solution of the differential equation: $$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$$
Step-by-Step Algorithmic Solution Trace:
- Find the first derivative of the given function $y = e^{-3x}$ with respect to $x$: $$\frac{dy}{dx} = -3e^{-3x}$$
- Find the second derivative by differentiating again: $$\frac{d^2y}{dx^2} = 9e^{-3x}$$
- Substitute $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ into the left-hand side (L.H.S.) of the differential equation: $$\text{L.H.S.} = 9e^{-3x} + (-3e^{-3x}) - 6(e^{-3x})$$
- Simplify the expression: $$\text{L.H.S.} = (9 - 3 - 6)e^{-3x} = 0 \cdot e^{-3x} = 0 = \text{R.H.S.}$$
Final Evaluation Output: Since L.H.S. = R.H.S., the function $y = e^{-3x}$ is verified as a solution.
We focus on three systematic classes of first-order, first-degree differential equations: variable separable, homogeneous, and linear equations.
If a first-order differential equation can be expressed in the form: $$\frac{dy}{dx} = g(x) \cdot h(y)$$ then it is of the variable separable type. We solve it by separating the variables and integrating both sides: $$\int \frac{1}{h(y)} dy = \int g(x) dx + C$$
Mathematical Problem Statement: Find the particular solution of the differential equation $\frac{dy}{dx} = -4xy^2$, given that $y = 1$ when $x = 0$.
Step-by-Step Algorithmic Solution Trace:
- Separate the variables $x$ and $y$: $$\frac{1}{y^2} dy = -4x dx$$
- Integrate both sides: $$\int y^{-2} dy = -4 \int x dx \implies -\frac{1}{y} = -2x^2 + C$$
- Express $y$ explicitly: $$y = \frac{1}{2x^2 - C}$$
- Substitute the initial conditions $x = 0, y = 1$ to find the particular constant $C$: $$1 = \frac{1}{2(0)^2 - C} \implies C = -1$$
- Substitute $C = -1$ back into the general solution: $$y = \frac{1}{2x^2 + 1}$$
Final Evaluation Output: Particular Solution: $y = \frac{1}{2x^2 + 1}$
A function $F(x, y)$ is a homogeneous function of degree $n$ if $F(\lambda x, \lambda y) = \lambda^n F(x, y)$ for any non-zero constant $\lambda$. A differential equation of the form $\frac{dy}{dx} = F(x, y)$ is homogeneous if $F(x, y)$ is a homogeneous function of degree zero.
Mathematical Problem Statement: Show that the differential equation $(x - y)\frac{dy}{dx} = x + 2y$ is homogeneous and solve it.
Step-by-Step Algorithmic Solution Trace:
- Rewrite the equation as: $$\frac{dy}{dx} = \frac{x + 2y}{x - y} = F(x, y)$$
- Verify homogeneity: $$F(\lambda x, \lambda y) = \frac{\lambda x + 2\lambda y}{\lambda x - \lambda y} = \frac{\lambda(x + 2y)}{\lambda(x - y)} = \lambda^0 F(x, y)$$ Thus, it is homogeneous of degree zero.
- Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$: $$v + x\frac{dv}{dx} = \frac{x + 2vx}{x - vx} = \frac{1 + 2v}{1 - v}$$
- Isolate the variable separable terms: $$x\\frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v = \frac{1 + 2v - v + v^2}{1 - v} = \frac{v^2 + v + 1}{1 - v}$$
- Separate variables and integrate: $$\int \frac{1 - v}{v^2 + v + 1} dv = \int \frac{dx}{x}$$
- Solve the integral on the left by splitting the numerator: $$- \frac{1}{2} \int \frac{2v + 1 - 3}{v^2 + v + 1} dv = \log |x| + C_1$$ $$- \frac{1}{2} \log |v^2 + v + 1| + \frac{3}{\sqrt{3}} \tan^{-1}\left(\frac{2v + 1}{\sqrt{3}}\right) = \log |x| + C_1$$
- Substitute $v = \frac{y}{x}$ back and simplify to obtain the general solution: $$\log |x^2 + xy + y^2| = 2\sqrt{3} \tan^{-1}\left(\frac{x + 2y}{\sqrt{3}x}\right) + C$$
Final Evaluation Output: General Solution: $\log |x^2 + xy + y^2| = 2\sqrt{3} \tan^{-1}\left(\frac{x + 2y}{\sqrt{3}x}\right) + C$
A first-order differential equation is linear if it can be written in one of the following two standard forms:
| Type | Standard Form | Integrating Factor (I.F.) | General Solution Formula |
|---|---|---|---|
| Type I | $$\frac{dy}{dx} + Py = Q$$ (where $P, Q$ are functions of $x$ only) |
$$\text{I.F.} = e^{\int P dx}$$ | $$y \cdot (\text{I.F.}) = \int (Q \cdot \text{I.F.}) dx + C$$ |
| Type II | $$\frac{dx}{dy} + P_1x = Q_1$$ (where $P_1, Q_1$ are functions of $y$ only) |
$$\text{I.F.} = e^{\int P_1 dy}$$ | $$x \cdot (\text{I.F.}) = \int (Q_1 \cdot \text{I.F.}) dy + C$$ |
Mathematical Problem Statement: Find the general solution of the differential equation: $$x\frac{dy}{dx} + 2y = x^2 \quad (x \neq 0)$$
Step-by-Step Algorithmic Solution Trace:
- Divide the entire equation by $x$ to convert it to standard linear form: $$\frac{dy}{dx} + \frac{2}{x}y = x$$ Here, $P = \frac{2}{x}$ and $Q = x$.
- Calculate the Integrating Factor (I.F.): $$\text{I.F.} = e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2\log x} = e^{\log(x^2)} = x^2$$
- Write the general solution using the formula: $$y \cdot (\text{I.F.}) = \int (Q \cdot \text{I.F.}) dx + C$$ $$y \cdot x^2 = \int (x \cdot x^2) dx + C$$
- Integrate the right-hand side: $$y x^2 = \int x^3 dx + C \implies y x^2 = \frac{x^4}{4} + C$$
- Solve for $y$: $$y = \frac{x^2}{4} + Cx^{-2}$$
Final Evaluation Output: General Solution: $y = \frac{x^2}{4} + Cx^{-2}$
Differential equations serve as the mathematical language of physical sciences, modeling rates of change across physics, chemistry, biology, and economics.
Historical Chronicle: Gottfried Wilhelm Leibniz (1646-1716)