First Order Differential Equation: Definition, Properties and Application

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An equation with two variables having one derivative is simply the first-order Differential equation. By equation, dy/dx =f (x,y) is an example of a first-order differential equation as it has two variables x and y, and one derivative dy/dx. As the name suggests, the equation must have only the first derivative dy/dx to be of the first order and no other higher-order derivative should exist in the same equation.

The first-order differential equation may be written as follows:

Y’ = f (x,y)

In general, the differential equation is used to represent a relationship between the function and its derivatives.

Table of Content

Writing the First Order Linear Differential Equation
Solutions of the First Order Linear Differential Equations
Properties of the First Order Linear Differential Equations
Applications of the First Order Linear Differential Equations
Things to Remember
Sample Questions

Key Terms: Differential equation, variable, Linear equation, intervals, linear differential equation, integrating factor

Writing the First Order Linear Differential Equation

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Now writing a first-order differential equation will look like,

Y’ = f (x,y)

Now that, this is the linear differential equation, it must be written in the form of,

Y’ + P(x)y = Q(x)

Where P and Q will be the functions of x and would exist in the same interval I.

Further, another way to write the first-order linear equation is,

(dy/dx) + P(x)y = Q (x)

Again where P(x) and Q(x) are the continuous functions of x.

Next, Let’s take a lookup at the ways of finding

Solutions of the First Order Linear Differential Equations

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Majorly there are two ways by which we find the solutions of first-order differential equations, namely they are:

Integrating Factor and
Method Of Variation Of Constant

Integrating Factor

In order to solve a first-order linear equation, first rewrite it in the standard form above, if it’s already in standard form, leave it like that; then multiply both sides by the integrating factor, which is

U(x) = ^e∫P.dx

After the multiplication of the left side of the integrating factor to the linear equation, the equation will look like this,

u.dy/dx + u.Py = u.Q,

Let’s take it as equation (1)

This formed equation is easy to solve further now, because the left side of the equation collapsed, let’s do it,

u.dy/dx + u.Py = u.dy/dx + ^e∫P.dx .Py

= u.dy/dx + y. d(^e∫P.dx)/dx

= u.dy/dx + y. du/dx

= d(u.dy)/dx

Therefore, equation (1) becomes

D(u.y)/dx = u.Q

Further, doing integration both sides, we will get,

u.y = ∫(u.Q) dx

Do not try to memorize the general solution instead learn the steps to get there.

Read Also: Methods of Integration

Method of Variation of a Constant

Just as the integrating method, we start with writing the equation in general form as,

Y’ + a(x)y = 0

Now, as we know that the general solution of a homogenous equation always contains an integration constant ‘C’.

In this method, we change the constant ‘C’ to an undetermined function C(x).

So when we would substitute the solution to the non-homogeneous equation, we will find the value of C(x).

The process is named as it works, the Method of Variation of a Constant.

Now, let’s look up at some of the properties of first-order differential equations.

Also Read:

Chapter Related Topics
Differential Equations Applications	Partial Differential Equation	Difference between parametric and nonparametric
Ordinary differential equations	Differential Calculus	Differential Equations Formula
Partial Derivative	Homogeneous Differential Equation	Homogeneous Functions

Properties of the First Order Linear Differential Equations

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Possesses no transcendental functions like trigonometric or logarithmic functions.
There’s no presence of the products of y and its existing derivatives.
Further, let’s discuss the applications of the first-order differential equations.

Read More: Applications of Derivatives

Applications of the First Order Linear Differential Equations

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The first-order differential equation has a wide scope of applications, some of them are named further:

Newton’s law of cooling
Falling Body Problems
Electrical circuits
Orthogonal trajectories
Growth and decay
Dilution Problems

Read More: Differentiation and Integration Formula

Things to Remember

A first-order differential equation is an equation with two variables having one derivative.
The equation must have only the first derivative dy/dx.
The equation can further be written in the following manner:

Y’ + P(x)y = Q(x)

or (dy/dx) + P(x)y = Q (x).

Integrating Factor and Method Of Variation Of Constant are the two main approaches for solving a first-order differential equation.

Read More: method of solving a differential equation

Sample Questions

Ques: Solve the provided DE (differential equation), Dx/dt – cos(t)x(t) = cos(t) (1) For the primary conditions of x(0)=0. (5 marks)

Ans: As it is a first-order linear differential equation, we can solve it by finding an integrating factor μ(t). Let’s take μ(t) to be,

Μ(t) = ^e−∫cos(t) = ^e−sin(t),

Now multiplying both sides of the door differential equation by μ, we can rewrite the differential equation as,

d/dt (^e−sin(t).x(t)) = ^e−sin(t).cos(t).

Next, Integrating both sides w.r.t. ‘t’, we get

^E−sin(t).x(t)=∫^e−sin(t).cos(t)dt + C

=^−e−sin(t) + C,

Next, we use the u-substitution (u=sin(t)) to calculate the required integral.

The general form of the solution to equation (1) is then calculated by dividing by ^e−sin(t).

X(t) = −1 + Ce^sin(t)

We double-check if the equation has a solution (1). Since

Dxdt=Cesin(t)cos(t)

We calculate that

Dxdt−cos(t)x(t)=Cesin(t)cos(t)−cos(t)(−1+Cesin(t))=cos(t),

Representing the found general solution to the differential equation.

At the end, we calculate the constant of integration ‘C’ by the condition

0 = x(0) = −1 + Ce⁰ = −1 + C,

So that C=1.

Our specific solution to the differential equation of (1) is

X(t) = −1 + ^esin(t).

Ques: Solve the further given differential equation Y’+2xy = x. (3 marks)

Ans: The differential equation presented above has already been transformed into a standard form.

Y’ + P(x)y = Q(x)

Therefore, P(x) = 2x and Q(x) = x

Ques: Solve the provided linear differential equation X².dy/dx = -2y (3 marks)

Ans: To start, first write the provided equation in the standard form,

Dy/dx + 2y/x² = 0

Let this equation be equation.(1)

Here, the integrating factor will be,

U(x) = ^e∫P.dx = ^e∫(2/x²). dx = e^-2/x

Now multiplying the equation (1) by μ = e^-2/x on both sides, we get

E^−2/x.dy/dx + 2.(e^−2/x).y /x² = 0

Next, collapsing the left-hand side, we get

D(e^−2/x.y) / dx = 0

And in the end, integrate:

E^−2/x.y = C

And the required answer is y = c.e^2/x

Ques: Solve the below given differential equation: Dy/dx – 4y/x = 0 (3 marks)

Ans: The given linear equation is already standard form, with P(x) = –4/ x. So,

∫P.dx = ∫-4/x.dx = -4 ln|x| = ln (x^-4)

Then, the integrating factor of the equation will be,

U(x) = ^e∫P.dx = e^ln (x^-4) = x^-4

Now multiplying by u(x) = x^-4 on both sides, we get,

X^-4.dy/dx – 4(x^-4).y /x = 0

Next, we will collapse the left side to get,

D((x^-4).y) /dx =0

And in the end, integrating the obtained equation, we will get our general solution,

X^-4.y = C

Or,

Y = Cx⁴

Ques: Solve the below given equation y′−y−x.e^x = 0 (3 marks)

Ans: Given equation, y′−y−x. e^x = 0

Rewriting the given equation, it becomes,

y′−y = x.e^x

Now by integrating factor, we get;

u(x) = ^e∫(−1)dx

u(x) = e−∫dx = e^(−x)

the general solution of the linear equation becomes,

y(x) = ∫u(x). f(x). dx +C / u(x)

y(x) = ∫e^(−x).xe^(x).dx +C / e^(−x)

y(x) = ∫x.dx +C / e^(−x)

= e^x. (x²/2 + C)

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First Order Differential Equation: Definition, Properties and Application

Writing the First Order Linear Differential Equation

Solutions of the First Order Linear Differential Equations

Properties of the First Order Linear Differential Equations

Applications of the First Order Linear Differential Equations

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
At a birthday party, children are being served orange juice in conical cups, as shown in the figure.

Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
On the basis of the above information, answer the following questions :

2.
Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

3.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

4.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

5.
Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

6.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

Comments

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First Order Differential Equation: Definition, Properties and Application

Writing the First Order Linear Differential Equation

Solutions of the First Order Linear Differential Equations

Properties of the First Order Linear Differential Equations

Applications of the First Order Linear Differential Equations

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.At a birthday party, children are being served orange juice in conical cups, as shown in the figure. Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.On the basis of the above information, answer the following questions :

2.Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

3. Find: The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

4.Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

5.Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

6. Find: If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
At a birthday party, children are being served orange juice in conical cups, as shown in the figure.

Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
On the basis of the above information, answer the following questions :

2.
Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

3.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

4.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

5.
Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

6.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).