Definitions
As we discussed earlier in here differential equations widely used when we need to model real-world actions. For instance, velocity, acceleration, heat on a surface etc. Remember that, derivatives measure how a quantity changes, so a differential equation describes a relationship between a quantity and its rate of change. Let us leave these intuitive definitions aside and focus on rigorous definitions of differential equations.
$\textbf{Definition 1.1.:} $ (Differential Equations) An equation containing the derivatives of one or more dependent variables with respect to one or more independent variables is said to be a differential equation, shortly DE.
$\textbf{Definition 1.2.:} $ (Ordinary Differential Equations) A differential equation that contains only ordinary derivatives of one or more dependent variables with respect to a single independent variable is called an ordinary differential equation, shortly ODE.
$$ \begin{aligned} \dfrac{dy}{dx} + 2y &= x^2y^2 \\ \dfrac{d^3y}{dx^3}-\dfrac{dy}{dx} + 5y &= 0 \end{aligned} $$
$\textbf{Definition 1.3.:} $ (Partial Differential Equations) A differential equation that contains partial derivatives of one or more dependent variables with respect to one or more independent variables is called a partial differential equation, namely PDE.
$$ \begin{aligned} \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} &= 0 \\ \dfrac{\partial u}{\partial y} &= \dfrac{\partial v}{\partial t} \end{aligned} $$
We have different writing styles for differential equations.
Leibniz Notation: $$ \begin{aligned} \dfrac{dy}{dx}, \dfrac{d^2y}{dx^2} \end{aligned} $$
Prime notation: $$ \begin{aligned} y^{\prime}, y^{\prime \prime}, y^{\prime \prime \prime} \end{aligned} $$
Newton’s Dot Notation: $$ \begin{aligned} \dot{s}, \ddot{s} \end{aligned} $$
Partial derivatives can be denoted by using subscript notation: $$ \begin{aligned} U_{xx}+U_{yy} = 4U_{xy} \end{aligned} $$
Classification
Classification by order
The order of an ODE or PDE is the order of the highest derivative in the equation. Consider the following ODE:
$$ \begin{align} \dfrac{d^2y}{dx^2} - 5\left(\dfrac{dy}{dx}\right)^4 +4y = \sin(x) \end{align} $$
Since the highest derivative in $(1)$ is $2$, we say the order of $(1)$ is $2$. We can write differential equations in different forms. In general, first-order ODE can be written as: $$ \begin{aligned} M(x,y) dx + N(x,y) dy = 0 \end{aligned} $$
And $n$-th order ODE can be written as follows: $$ \begin{aligned} F(x, y, y^{\prime}, y^{\prime \prime},\dots, y^{(n)}) = 0 \end{aligned} $$
Here $x$ is independent variable and the rest $y$s are dependent variables.
Classification by linearity
A DE is classified as linear or nonlinear by how the dependent variable(s) and their derivatives appear.
- Linear differential equation: The dependent variable(s) and all their derivatives should appear only to the first power, should not be multiplied by each other and should not be inside nonlinear functions line $\sin(y)$, $\exp(y)$, etc.
For a single-function ODE, a general linear form is written as follows:
$$ \begin{aligned} a_n(x)y^{(n)} +\dots+a_0(x)y = g(x) \end{aligned} $$
where $a_i(x)$ and $g(x)$ are given functions of the independent variable only.
- Nonlinear differential equation: Any DE that violates the conditions above is nonlinear.
$$ \begin{aligned} \left(\dfrac{dy}{dx}\right)^4 + 4y = \sin(x) \end{aligned} $$
is nonlinear because $\left(\dfrac{dy}{dx}\right)^4$.
Yes. Not because $\sin(x)$, since $x$ is independent variable and thus does not violates the conditions on linearity. Consider the following ODE:
$$ \begin{aligned} \dfrac{d^2y}{dx^2} + \alpha \sin(y) = 0 \end{aligned} $$
It is nonlinear because the term $\sin(y)$ violates the conditions of linearity. I suggest strong comprehension for these topics. Moving on to other sections without comprehending it very well leads to a breakdown in the chain and a lack of understanding of what is being done. If you don’t understand, please reread or try another source before continuing.
Now that we have given the essential definitions, we can use them in theorems and questions.
Solution of DEs
A solution of the ODE on the interval $\alpha < x < \beta$ is a function $y = \varphi(x)$ such that $\varphi(x)^{\prime}$, $\varphi(x)^{\prime\prime}$, $\dots$, $\varphi(x)^{n-1}$ exist and satisfy $$y^{(n)}=f(x, y’, y’’, \dots, y^{(n-1)})$$ for every $\alpha < x <\beta$.
We may refer to family of solutions by saying solution of DE. Some DEs have no solution for given data, some have several, some infinitely many. For example, the differential equation $\dfrac{d^2y}{dx^2}+4y=0$ has infinitely many solutions, namely $y(x)=c_1\cos(2x)+c_2\sin(2x)$, where $c_1,c_2\in\mathbb{R}$.
You might think the derivation process of the solution. Well, we will generate formulas for famous differential equations and will find generalized solutions. Let us start with the easist.
Separable DEs
If the differential equation is in the form of $$ \begin{aligned} \dfrac{dy}{dx} = f(x,y) \end{aligned} $$
or
$$ \begin{align} M(x,y)dx + N(x,y)dy=0 \end{align} $$
moreover if $M(x,y) = M(x)$ and $N(x,y)=N(y)$ then
$$ \begin{align} M(x)dx + N(y)dy=0 \end{align} $$
The forms in $(2)$ and $(3)$ are said to be separable differential equation. To solve we integrate the both sides of $(3)$ as follows:
$$ \begin{aligned} \int M(x,y)dx + \int N(x,y)dy = 0 \end{aligned} $$
and we get $\mathcal{H}_1(x,y)+\mathcal{H}_2(x,y)=C$ as solution of $(3)$, where $C$ is an integration constant.
Common real-life separable equations are population growth e.g. $\dfrac{dP}{dt} = kP$, radioactive decay with equation $\dfrac{dA}{dt} = -kA$ and so on.
$\textbf{Example:}$ Find the general solution of the given nonlinear differential equation.
$$ \begin{align} \dfrac{dy}{dx} = \dfrac{2x}{2y+1} \cdot \dfrac{1+y^2}{e^{-x^2}} \end{align} $$
$\textbf{Solution:}$ Let us separate the terms according to the type of differential equation, as mentioned.
$$ \begin{aligned} \dfrac{2y+1}{1+y^2}dy &= \dfrac{2x}{e^{-x^2s}}dx \\ \dfrac{2y}{1+y^2}dy + \dfrac{1}{1+y^2}dy &= \dfrac{2x}{e^{-x^2}}dx\\ \end{aligned} $$
And integrating both sides with respect to $x$ and $y$ we obtain
$$ \begin{align} & \int \dfrac{2y}{1+y^2}dy + \int \dfrac{1}{1+y^2}dy \notag \\ = &\int \dfrac{2x}{e^{-x^2}}dx \\ \end{align} $$
Integrating the terms would give us: $$ \begin{align} & \ln(1+y^2)+\arctan(y) \notag \\ = &e^{x^2}+C \end{align} $$
$(6)$ is the integral curves or implicit form of the solution of the differential equation $(4)$ where $C$ is the constant of integration. The graph below belongs to $(6)$ for $C=1$. One can also create such graphs by using Desmos
As $C$ differs, the graph changes and so solution. To fix $C$ we have a real big problem set called initial value problems, shortly IVP.
Homogenous DEs
If the differential equation is in the form of $$ \begin{align} \dfrac{dy}{dx}=F(x,y)=F\left(\dfrac{x}{y}\right) \end{align} $$
or if we let $y=ux$ in $M(x,y)dx+N(x,y)dy=0$ $$ \begin{aligned} &\Rightarrow M(x,ux)dx+N(x,ux)dy \notag\\ &=x^\alpha(M(1,u)dx+N(1,u)dy)\notag\\ &=0 \end{aligned} $$
is said to be homogenous differential equation with the degree of $\alpha$. Then the substitution $\dfrac{y}{x}=u$ transforms homogenous differential equation into separable differential equation. Consider the equation $(7)$ and use the transform $y = ux$ $$ \begin{aligned} \dfrac{dy}{dx} = F\left(\dfrac{y}{x}\right) \end{aligned} $$
We have
$$ \begin{align} &\dfrac{dy}{dx} = \dfrac{d(u(x))}{dx} \notag \\ &= u + x\cdot u^{\prime} \notag \\ &= F(u) \end{align} $$
In fact, we have a separable equation $$ \begin{align} u^{\prime} \cdot x = F(u) - u \end{align} $$
or similarly, $$ \begin{align} \dfrac{du}{F(u)-u} = \dfrac{dx}{x} \end{align} $$
On $(10)$ we integrate the both sides with respect to $u$ and $x$ $$ \begin{align} \int \dfrac{du}{F(u)-u} = \int \dfrac{dx}{x} \end{align} $$
The solution to $(11)$ is $$ \begin{align} \vert Cx \vert = e^{\displaystyle \int \frac{du}{F(u)-u}} \end{align} $$
And by back substituting $\dfrac{y}{x} = u$ in $(12)$ we obtain the general solution of given homogenous differential equation. Observe that homogenous differential equations does not care about scale, it only takes ratio $\frac{y}{x}$ into account. In real-life we see these kind of equations when scaling all lengths by the same factor doesn’t change the law. For instance, direction fields in central-force problems, certain geometric optics or ray‑tracing problems and etc.
$\textbf{Example:}$ Use an appropriate substitution to solve the given differential equation subject to the indicated initial condition $$ \begin{aligned} x\cdot \dfrac{dy}{dx}-y = \dfrac{x^2}{y}e^{-\dfrac{y}{x}} \end{aligned} $$ where $y(1) = 2$
$\textbf{Solution:}$ We set $$ \begin{align} \dfrac{dy}{dx} = \dfrac{y}{x} + \dfrac{x}{y}e^{-\dfrac{y}{x}} = F\left(\dfrac{y}{x}\right) \end{align} $$
So that we have a homogenous differential equation. Now, recall the transform $y=ux$ and its differential $$ \begin{align} y^{\prime} &= u + x \cdot u^{\prime} \end{align} $$
Using $(14)$ in the equation $(13)$ would give us a separable differential equation $$ \begin{align} x\cdot u^{\prime} + u = u + \dfrac{1}{u}e^{-u} \end{align} $$
Rearranging the equation $(15)$ $$ \begin{align} x \cdot u^{\prime} &= \dfrac{e^{-u}}{u} \notag \\ u \cdot e^{u}du &= \dfrac{dx}{x} \end{align} $$
Integrating both sides with respect to $u$ and $x$. We obtain general solution $$ \begin{align} \overbrace{\int u \cdot e^{u}du}^{\text{(1)}} = \overbrace{\int \dfrac{dx}{x}}^{\text{(2)}} \end{align} $$
To integrate $(17.1)$ we use integration by parts (e.g. use $u = t$ and $e^u = w$ and $tw - \int wdt$). For $(17.2)$ it is trivial. Gathering the solutions we obtain: $$ \begin{aligned} \left(u - 1\right)e^{u} &=\ln{\vert C\cdot x\vert} \end{aligned} $$
Back substituting $u = \dfrac{y}{x}$ $$ \begin{aligned} \left(\dfrac{y}{x} - 1\right)e^{\dfrac{y}{x}} &= \ln{\vert C\cdot x\vert} \end{aligned} $$
Now, we use the initial condition $y(1) = 2$, $$ \begin{align} \left( \dfrac{2}{1} - 1\right)e^{2} &= \ln{\vert C \cdot 1 \vert} \notag \\ e^2 &= \ln{\vert C \vert} \notag \\ e^{e^{2}} &= C \end{align} $$
Hence, the solution of the initial value problem is $$ \begin{align} (y-x)e^{\dfrac{y}{x}} = x(\ln{\vert e^{e^2}\cdot x\vert}) \end{align} $$
In the next section, we will deep dive into first order differential equations, substitution methods, linear differential equations, Bernoulli’s differential equation and Ricatti’s differential equations.