First Order Differential Equations and its types

First Order Differential Equations and Its Types: A first order differential equation is an equation that relates the derivative of an unknown function to the function itself. First order differential equation is an eqtn that involves only the 1st derivative of the function which is unknown. The general form of a first-order ordinary differential equation (ODE) is:

$\frac{{dy}}{{dx}} = f(x, y)$

where $\frac{{dy}}{{dx}}$ represents the derivative of the unknown function $y$ with respect to the independent variable $x$ , and $f(x, y)$ is a given function of both $x$ and $y$ .

The method of solving a particular differential equation depends on its type and can involve techniques such as separation of variables, integrating factors, substitutions, or using specific solution methods for certain types of equations.

First-order differential equations can be classified into different types, depending on the nature of the function $f(x, y)$ . Some common types of first-order ODEs include:

Example 1:

Solve the differential equation

$\frac{dy}{dx} = 2x.$

Solution:
To solve this equation, we can separate the variables and integrate both sides with respect to $x$ :

${dy} ={2x}{dx}$

Integrating both sides:

$\int dy = \int {2x}{dx}$

$y = x^2 + C$

where $C$ is the constant of integration.

Example 2:

Solve the differential equation

$\frac{dy}{dx} = 3y.$

. Solution:
This equation is a separable differential equation. We can separate the variables and integrate as follows:

$\frac{dy}{y} = 3dx$

Integrating both sides:

$\int \frac{dy}{y} = \int 3dx$

$\ln|y| = 3x + C$

Taking the exponential of both sides:

$|y| = e^{3x+C}$

Since the absolute value can be positive or negative, we can rewrite it as:

$y = \pm e^{3x+C}$

where $\pm$ indicates that the solution can take either the positive or negative form. We can simplify the expression further by combining the constant $C$ and the sign:

$y = Ke^{3x}$

where $K$ is a nonzero constant.

Initial Value Problem

Initial Points (IP)

The answer (or its derivatives) must fulfil the initial condition(s), which is/are a collection of points.

Initial Conditions

Examples:
For a (DE) equation which involved in a function f(t). the initial conditions (ICs) are of the form: $f(t_0) = f_0, f'(t_0) = f'_0, f''((t_0)=f''_0,..., \quad etc$

Initial Value Problem.

A differential equation having initial conditions is an initial value problem, or IVP for short.

Separable Differential Equations

In this type, the equation can be rearranged in a way that separates the variables $x$ and $y$ , allowing them to be integrated separately. The general form is

$\frac{{dy}}{{dx}} = g(x)h(y)$

, where $g(x)$ and $h(y)$ are functions of $x$ and $y$ , respectively.

Question # 1:

$\frac{d y}{d x}=\frac{x^2}{y\left(1+x^3\right)}$

[expand title=”Step by Step Solution“]

$\begin{aligned} & \frac{d y}{d x}=\frac{x^2}{y\left(1+x^3\right)} \\ & y d y=\frac{x^2}{1+x^3} d x \\ & \int y d y=\frac{1}{3} \int \frac{3 x^2}{1+x^3} \cdot d x \\ & \frac{y^2}{2}=\frac{1}{3} \ln \left(1+x^3\right)+c \\ & \frac{3 y^2}{2}=\ln \left(1+x^3\right)+3 c \\ & 3 y^2=2 \ln \left(1+x^3\right)+6 c \\ & 3 y^{2}=2 \ln \left(1+x^3\right)+c^{\prime} \\ & \end{aligned}$

[/expand]

Question # 2:

$\frac{d y}{d x}+y^{2} \sin x=0$

[expand title=”Step by Step Solution“] Given equation is

$\begin{aligned} &\frac{d y}{d x}+y^{2} \sin x=0\\ &\Rightarrow \frac{d y}{d x}=-y^{2} \sin x\\ &\Rightarrow \frac{d y}{y^{2}}=-\sin x d x\\$

Integrating both sides

$\begin{aligned} &\int \frac{d y}{y^{2}}=-\int \sin x d x\\ &\Rightarrow \frac{-1}{y}=-(-\cos x)+c\\ &\Rightarrow \frac{1}{\mathrm{y}}+\cos \mathrm{x}=\mathrm{c} \end{aligned}$

is required solution. [/expand]

Question # 3:

$\frac{d y}{d x}=1+x+y^{2}+x y^{2}$

[expand title=”Step by Step Solution“] Given equation is

$\begin{aligned} &\frac{d y}{d x}=1+x+y^{2}+x y^{2}\\ &\frac{d y}{d x}=1(1+x)+y^{2}(1+x)\\ &\frac{d y}{d x}=\left(1+y^{2}\right)(1+\mathrm{x})\\ & \frac{d y}{1+y^{2}}=(1+\mathrm{x}) \mathrm{dx}\\ &\int \frac{\mathrm{dy}}{1+y^{2}}=\int(1+x) d x\\ & \tan ^{-1}(y)=\int d x+\int x d x\\ & \tan ^{-1}(\mathrm{y})-\mathrm{x}-\frac{\mathrm{x}^{2}}{2}=\mathrm{c}$ \end{aligned}$

is required solution. [/expand]

Question # 4:

$(x y+2 x+y+2) d x+\left(x^{2}+2 x\right) d y=0$

[expand title=”Step by Step Solution“] Given equation is

$\begin{aligned} &(x y+2 x+y+2) d x+\left(x^{2}+2 x\right) d y=0\\ &(x y+2 x+y+2) dx=-\left(x^{2}+2 x\right) dy\\ &\mathrm{y}(\mathrm{x}+1)+2(\mathrm{x}+1) \mathrm{dx}=-\left(\mathrm{x}^{2}+2 \mathrm{x}\right)\mathrm{dy}\\ &(\mathrm{y}+2)(\mathrm{x}+1) \mathrm{dx}=-\left(\mathrm{x}^{2}+2 \mathrm{x}\right) \mathrm{dy}\\ & \frac{d y}{y+2}=-\frac{x+1}{\left(x^{2}+2 x\right)} d x\\ &\int \frac{d y}{y+2}=-\int \frac{x+1}{\left(x^{2}+2 x\right)} d x\\ & \int \frac{d y}{y+2}=-\frac{1}{2} \int \frac{2 x+2}{x^{2}+2 x} d x\\ & \ln (\mathrm{y}+2)=-\frac{1}{2} \ln \left(\mathrm{x}^{2}+2 \mathrm{x}\right)+\mathrm{c} \end{aligned}$

is required solution. [/expand]

Question # 5:

$\frac{d y}{d x}=2 x^{2}+y+x^{2} y+x y-2 x-2$

[expand title=”Step by Step Solution“] Given equation is

$\begin{aligned} &\frac{d y}{d x}=2 x^{2}+y+x^{2} y+x y-2 x-2\\ & \frac{d y}{d x}=y\left(1-x^{2}+x\right)+2\left(x^{2}-x-1\right)\\ & \frac{d y}{d x}=y\left(1-x^{2}+x\right)-2\left(1-x^{2}+x\right)\\ & \frac{\mathrm{dy}}{\mathrm{dx}}=(\mathrm{y}-2)\left(1-x^{2}+\mathrm{x}\right)\\ & \frac{d y}{y-2}=\left(1-x^{2}+x\right) d x\\ & \int \frac{d y}{y-2}=\int\left(1-x^{2}+x\right) d x\\ & \int \frac{d y}{y-2}=\int d x-\int x^{2} d x+\int x d x & \ln (\mathrm{y}-2)=\mathrm{x}-\frac{\mathrm{x}^{3}}{3}+\frac{\mathrm{x}^{2}}{2}+\mathrm{c}_{1}\\ & 6 \ln (\mathrm{y}-2)=6 \mathrm{x}-2 \mathrm{x}^{3}+3 \mathrm{x}^{2}+6 \mathrm{c}_{1}\\ &-6 \ln (\mathrm{y}-2)=-6 x+2 \mathrm{x}^{3}-3 \mathrm{x}^{2}-6 \mathrm{c}_{1}\\ & \ln (\mathrm{y}-2)^{-6}=2 \mathrm{x}^{3}-3 \mathrm{x}^{2}-6 \mathrm{x}+\mathrm{c}\\ &(\mathrm{y}-2)^{-6}=\mathrm{e}^{2 \mathrm{x}^{3}-3 \mathrm{x}^{2}-6 \mathrm{x}+\mathrm{c}}\\ \end{aligned}$

is required solution. [/expand]

Question # 6:

$\operatorname{cosec} y d x+\sec x d y=0$

[expand title=”Step by Step Solution“] Given equation is

$\begin{aligned} &\operatorname{cosec} y d x+\sec x d y=0\\ & \operatorname{cosecydx}=-\operatorname{cosecy} d x\\ & \frac{d x}{\sec x}=-\frac{d y}{\operatorname{cosec} y}\\ & \int \frac{\mathrm{dx}}{\sec x}=-\int \frac{\mathrm{dy}}{\operatorname{cosecy}}\\ & \int \cos x d x=-\int \sin y d y\\ & \sin \mathrm{x}=-(-\cos \mathrm{x})\\ & \sin x=\cos x+c\\ & \sin \mathrm{x}-\cos \mathrm{x}=\mathrm{c}\\ \end{aligned}$

is required solution. [/expand]

Question # 7:

$y(1+x) d x+x(1+y) d y$

[expand title=”Step by Step Solution“] Given equation is

$\begin{aligned} &y(1+x) d x+x(1+y) d y\\ & \mathrm{y}(1+\mathrm{x}) \mathrm{d} x=-\mathrm{x}(1+\mathrm{y}) \mathrm{dy}\\ & \frac{(1+x)}{x} d x=-\frac{(1+y)}{y} d y\\ & \int \frac{(1+\mathrm{x})}{\mathrm{x}} \mathrm{dx}=-\int \frac{(1+\mathrm{y})}{\mathrm{y}} \mathrm{dy}\\ & \int \frac{d x}{x}+\int \frac{x}{x} d x=-\int \frac{d y}{y}-\int \frac{y}{y} d y\\ & \ln \mathrm{x}+\mathrm{x}=-\ln y-\mathrm{y}+\mathrm{c}\\ & \mathrm{x}+\mathrm{y}+\ln \mathrm{x}+\ln \mathrm{y}=\mathrm{c}\\ & \mathrm{x}+\mathrm{y}+\ln |\mathrm{xy}|=\mathrm{c} \end{aligned}$

is required solution. [/expand]

Question # 8:

$y \sqrt{1+x^{2}} d x+x \sqrt{1+y^{2}} d y=0$

[expand title=”Step by Step Solution“] Given equation is

$\begin{aligned} &y \sqrt{1+x^{2}} d x+x \sqrt{1+y^{2}} d y=0\\ & \mathrm{y} \sqrt{1+\mathrm{x}^{2}} \mathrm{dx}=-\mathrm{x} \sqrt{1+\mathrm{y}^{2}} \mathrm{dy}\\ & \frac{\sqrt{1+\mathrm{x}^{2}}}{\mathrm{x}} \mathrm{dx}=-\frac{\sqrt{1+\mathrm{y}^{2}}}{\mathrm{y}} \mathrm{dy}\\ & \int \frac{\sqrt{1+\mathrm{x}^{2}}}{\mathrm{x}} \mathrm{dx}=-\int \frac{\sqrt{1+\mathrm{y}^{2}}}{\mathrm{y}} \mathrm{dy}---\\$

Consider

$\mathrm{I}_{1}=\int \frac{\sqrt{1+\mathrm{x}^{2}}}{\mathrm{x}} \mathrm{dx}\\ \begin{gathered} \text { put } x=\tan \theta \\ dx=\sec ^{2} \theta \end{gathered}$

Therefore,

$\begin{aligned} &I_{1}=\int \frac{\sec \theta \cdot \sec ^{2} \theta}{\tan \theta} \\ & I_{1}=\int \frac{\sec \theta \cdot\left(1+\tan ^{2} \theta\right) \mathrm{d} \theta}{\tan \theta} \\ & I_{1}=\int \frac{\sec \theta}{\tan \theta} \mathrm{d} \theta+\int \frac{\sec \theta \cdot \tan ^{2} \theta}{\tan \theta} \mathrm{d} \theta \\ & I_{1}=\int \frac{1 / \cos \theta}{\sin \theta / \cos \theta} d \theta+\int \sec \theta \cdot \tan \theta d \theta \\ & I_{1}=\int \frac{1}{\sin \theta} \mathrm{d} \theta+\int \sec \theta \cdot \tan \theta \mathrm{d} \theta \\ & I_{1}=\int \operatorname{cosec} \theta \mathrm{d} \theta+\int \sec \theta \cdot \tan \theta \mathrm{d} \theta \\ & I_{1}=\ln (\operatorname{cosec} \theta-\cot \theta)+\sec \theta \\ & x=\tan \theta \\ & \cot \theta=\frac{1}{\mathrm{x}} \\ & \operatorname{cosec} \theta=\sqrt{\cot ^{2} \theta+1} \\ & \operatorname{cosec} \theta=\sqrt{\frac{1}{\mathrm{x}^{2}}+1} \\ & \operatorname{cosec} \theta=\frac{\sqrt{1+x^{2}}}{x} \end{aligned}$

Therefore,

$&I_{1}=\ln \left(\frac{\sqrt{1+x^{2}}}{x}-\frac{1}{x}\right)+\sec \theta\\ & \int \frac{\sqrt{1+x^{2}}}{x} d x=\ln \left(\frac{\sqrt{1+x^{2}}-1}{x}\right)+\sqrt{1+x^{2}}$

Similarly,

$\int \frac{\sqrt{1+y^{2}}}{\mathrm{y}} d y=\ln \left(\frac{\sqrt{1+y^{2}}-1}{\mathrm{y}}\right)+\sqrt{1+y^{2}}$

Therefore, (1) becomes

$\begin{aligned} & \ln \left(\frac{\sqrt{1+x^{2}}-1}{x}\right)+\sqrt{1+x^{2}} \\ & =-\ln \left(\frac{\sqrt{1+\mathrm{y}^{2}}-1}{\mathrm{y}}\right)-\sqrt{1+\mathrm{y}^{2}} \\ & +\mathrm{c} \\ & \sqrt{1+\mathrm{x}^{2}}+\sqrt{1+\mathrm{y}^{2}} \\ & +\ln \left(\frac{\left(\sqrt{1+x^{2}}-1\right)\left(\sqrt{1+y^{2}}-1\right)}{x y}\right) \\ & =\mathrm{c}\end{aligned}$

is required solution. [/expand]

Linear Differential Equations

These equations have the form

$\frac{{dy}}{{dx}} + p(x)y = q(x)$

, where $p(x)$ and $q(x)$ are given functions of $x$ . Linear differential equations can be solved using integrating factors or other methods.

Homogeneous Differential Equations

In this type, the equation can be expressed in the form

$\frac{{dy}}{{dx}} = F(y/x)$

, where $F$ is a function of the ratio $y/x$ . Homogeneous equations can be solved using a substitution such as $y = vx$ .

Exact Differential Equations

These equations can be written in the form

$M(x, y)dx + N(x, y)dy = 0$

, where $M$ and $N$ are functions of $x$ and $y$ . Exact equations can be solved by finding a function called the potential function or integrating factor.

Bernoulli Differential Equations

These equations have the form

$\frac{{dy}}{{dx}} + p(x)y = q(x)y^n$

, where $p(x)$ and $q(x)$ are given functions of $x$ , and $n$ is a constant. Bernoulli equations can be transformed into linear equations by making an appropriate substitution.

Initial Value Problem

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