Study Notes on Differential Equations of First Order and Higher Degree

Fundamentals of Differential Equations

Equations that are composed of an unknown function and its derivatives are formally called differential equations. These equations play a fundamental role in the field of engineering because many physical phenomena are best formulated mathematically in terms of their rate of change. A primary example is Newton’s second law with drag, where the dynamics are described by variables and constants such as $v$ (velocity, the dependent variable), $t$ (time, the independent variable), $m$ (mass), $g$ (acceleration due to gravity), and $c$ (drag co-efficient). In this context, $m$, $g$, and $c$ function as constants.

Ordinary and Partial Differential Equations

When a function involves only one independent variable, the equation is identified as an ordinary differential equation (or ODE). Examples of ODEs include expressions where $x$ is the independent variable and $y$ is the dependent variable, such as $\frac{dy}{dx} = 5x + 3$, $\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + 2y = 0$, and $(\frac{dy}{dx})^2 + y = ext{sin}(x)$. Conversely, a partial differential equation (or PDE) involves two or more independent variables. For instance, in the equation $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = u$, the variable $u$ is the dependent variable while $x$ and $y$ are independent. Similarly, in the wave equation $\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}$, $u$ is the dependent variable and $x$ and $t$ are the independent variables.

Order and Degree of Differential Equations

The order of a differential equation is defined as the order of the highest derivative present in the equation. Differential equations are classified accordingly: a first-order equation includes a first derivative as its highest derivative, a second-order equation includes a second derivative, and so on. Examples of order include $\frac{dy}{dx} = e^x$ (Order 1), $\frac{d^2y}{dx^2} + y = 0$ (Order 2), and $(\frac{d^3y}{dx^3})^2 + (\frac{d^2y}{dx^2})^5 + \frac{dy}{dx} = e^x$ (Order 3). The degree of a differential equation is the power of the highest order derivative term in the equation. For example, $\frac{dy}{dx} = e^x$ is of Degree 1, $\frac{d^2y}{dx^2} + y = 0$ is of Degree 1, and $(\frac{d^3y}{dx^3})^3 + (\frac{d^2y}{dx^2})^5 + \frac{dy}{dx} = e^x$ is of Degree 3.

Linearity of Differential Equations

A differential equation is considered linear if it satisfies two specific criteria: (1) every dependent variable and its derivatives are of degree one, and (2) no product of dependent variables and/or their derivatives occurs. For example, $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = 0$ is linear. However, $\frac{d^2y}{dx^2} + 2(\frac{dy}{dx})^2 + y = 0$ is non-linear because the second term is not of degree one. The equation $\frac{d^2y}{dx^2} + y\frac{dy}{dx} = e^x$ is non-linear because the second term involves the product of $y$ and $\frac{dy}{dx}$. Furthermore, $\frac{d^2y}{dx^2} + \text{sin}(y) = 0$ is non-linear because $\text{sin}(y)$ is a non-linear term.

Classification and Characteristics Table

Differential equations are categorized based on their type, linearity, variables, order, and degree. For the equation $y' = x + 6y$, it is an Ordinary, Linear equation of Order 1 and Degree 1, with independent variable $x$ and dependent variable $y$. The equation $y'' = 4y + y^3$ is Ordinary, Nonlinear, Order 2, Degree 1, with variables $x$ and $y$. The equation $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = x^3$ is Ordinary, Linear, Order 2, Degree 1. The equation $y'' + 2xy' + 4y = \text{cos}(2x)$ is Ordinary, Linear, Order 2, Degree 1. The equation $\frac{d^3y}{dx^3} + y + 4 = 0$ is Ordinary, Linear, Order 3, Degree 1. For partial differential equations, $\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0$ is Partial, Linear, Order 1, Degree 1, with independent variables $x, y, z$ and dependent variables $u, v, w$. The equation $\frac{\partial^2 u}{\partial x^2} = c^2 \frac{\partial^2 u}{\partial t^2}$ is Partial, Linear, Order 2, Degree 1, with independent variables $x, t$ and dependent variable $u$.

Solution Types and Boundary Conditions

Differential equations can be identified by further characteristics such as homogeneity and boundary conditions. Examples include structures that are 1st order, linear, nonhomogeneous, and part of an Initial Value Problem (IVP). Others may be 2nd order, linear, nonhomogeneous, and characterized as a Boundary Value Problem (BVP). Some equations are 2nd order, linear, and homogeneous (IVP), while others are 2nd order, nonlinear, and homogeneous (IVP). A solution or integral of a differential equation is any relation between the dependent and independent variables that reduces the equation to an identity when substituted. For example, $y = ce^{2x}$ is a solution of $y' = 2y$, where $c$ is a constant. Because this solution is true for all values of $c$, it is termed a general solution. When the arbitrary constant takes a unique value, it becomes a particular solution.

Methods for Solving First Order First Degree Equations

There are four primary methods for solving first order and first degree equations. The first is Separation of Variables, applied to equations in the form $f(y)dy = g(x)dx$. The method involves converting the equation to standard form and integrating both sides. For the example $(x+1)\frac{dy}{dx} = x(y^2+1)$, the solution is $\text{tan}^{-1}(y) = x - \log(1 + x) + c$. The second method involves Homogeneous Differential Equations, where $\frac{dy}{dx} = \frac{f(x,y)}{g(x,y)}$ and every term in $f$ and $g$ possesses the same degree. The procedure is to put $y = vx$, yielding a variable-separable form in $v$ and $x$. For $(2xy + x^2) \frac{dy}{dx} = 3y^2 + 2xy$, the solution is $y^2 + xy = cx^3$.

The third method targets Linear Differential Equations of the form $\frac{dy}{dx} + Py = Q$, where $P$ and $Q$ are functions of $x$ or constants. The method requires multiplying by an integrating factor $I.F. = e^{\int P\,dx}$ and integrating. For $(x + 1) \frac{dy}{dx} = y + e^x (x + 1)^2$, the solution is $\frac{y}{x+1} = e^x + c$. The fourth method is for Exact Differential Equations, which take the form $Mdx + Ndy = 0$ and satisfy $M_y = N_x$. The solution involves integrating $M$ with respect to $x$ (keeping $y$ constant), integrating $x$-deficient terms of $N$ with respect to $y$, and setting the sum to a constant. For $(5x^4 + 3x^2y^2 - 2xy^3)dx + (2x^3y - 3x^2y^2 - 5y^4)dy = 0$, the solution is $x^5 + x^3y^2 - x^2y^3 - y^5 = c$.

Differential Equations of First Order and Higher Degree

This section addresses the set of all first order higher degree ODEs, categorizing them into several solvable types: (1) Equations solvable for $p$ (where $p = \frac{dy}{dx}$), (2) Equations solvable for $y$, (3) Equations solvable for $x$, (4) Equations homogeneous in $x$ and $y$, (5) Clairaut equation, and (6) Lagrange equation. For equations solvable for $p$, with general form $f(x,y,p) = 0$, the goal is to solve for $p$ first and then integrate. In Example 1, solving $x^2 = 1 + p^2$ leads to $p = \pm \sqrt{x^2 - 1}$. Integration results in the solution $y = \pm \frac{x}{2}\sqrt{x^2 - 1} + \frac{1}{2}\log(x + \sqrt{x^2 - 1}) + c$.

Equations Solvable for y and x

For equations solvable for $y$, the method involves differentiating the given ODE with respect to $x$, eliminating $p$ between the original and the new equation, and identifying the eliminant as the solution. In Example 2, solving $y = (x - a)p - p^2$ involves differentiating to get $\frac{dy}{dx} = p + (x - a)\frac{dp}{dx} - 2p\frac{dp}{dx}$. Simplification yields $0 = (x - a - 2p)\frac{dp}{dx}$, which implies $\frac{dp}{dx} = 0$ and thus $p = c$. Substituting $p = c$ back into the ODE gives $y = (x - a)c - c^2$. For equations solvable for $x$, the method is to write the ODE in terms of $x$, differentiate with respect to $y$, solve the resulting equation, and eliminate $p$. In Example 3, solving $y = 2px + yp^2$ (rearranged as $2x = \frac{y}{p} - yp$) involves differentiating with respect to $y$ to eventually find $py = c$. Substituting $p = \frac{c}{y}$ into the ODE results in $y^2 = c(2x + c)$.

Clairaut’s Equation and Practice Problems

Clairaut’s Equation has the general form $y = px + f(p)$. Differentiating with respect to $x$ yields $p = p + x\frac{dp}{dx} + f'(p)\frac{dp}{dx}$, leading to $(x + f'(p))\frac{dp}{dx} = 0$. This implies $\frac{dp}{dx} = 0$, so $p = a$, where $a$ is a constant. The required solution is obtained by replacing $p$ with $a$: $y = ax + f(a)$. Example 4 demonstrates this with $p = \log(px - y)$. Rearranging gives $e^p = px - y$, or $y = px - e^p$. The solution is $y = ax - e^a$.

Unsolved examples for practice include:

1. $xp^2 + x = 2yp$ with the answer $2cy = c^2x^2 + 1$.
2. $x(1 + p^2) = 1$ with the answer $y - c = \sqrt{x - x^2} - \text{tan}^{-1}(\sqrt{\frac{1-x}{x}})$.
3. $x^2p^2 + xyp - 6y^2 = 0$ with answers $y = x^3$ and $y = C_1x^2$.
4. $y = px + p^3$ with the answer $y = ax + a^3$.