Differential Equations - Topics, Types, Books, FAQs

Differential Equations

A differential equation is any equation that contains at least one derivative of an unknown function. In class 12 differential equations, these equations describe how quantities change with respect to each other and are widely used in physics, engineering, biology, and economics.

General Form of Differential Equations

The general form of a differential equation is: $\frac{dy}{dx} = f(x)$

Here, $x$ is the independent variable, and $y$ is the dependent variable. Learning how to solve differential equations involves finding the function $y$ that satisfies the given relation.

Examples of Differential Equations

First-order differential equation: $x \frac{dy}{dx} + 2y = 0$
This equation involves the derivative of the dependent variable $y$ with respect to the independent variable $x$, making it a differential equation.

• $\frac{dy}{dx} = \sin 2x + \cos x$

• $k \frac{d^2y}{dx^2} = \left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}$

These examples illustrate different types of ordinary differential equations (ODEs) and partial differential equations (PDEs).

Order of Differential Equations

The order of a differential equation is defined as the highest order derivative present in the equation.

Examples:
$ \frac{dy}{dx} = \sin 2x + \cos x \quad \text{(First order)} $
$ \frac{d^2y}{dx^2} + y = 0 \quad \text{(Second order)} $
$ \frac{d^3y}{dx^3} + 2\left(\frac{d^2y}{dx^2}\right)^5 + y = e^x \quad \text{(Third order)} $

Types of differential equations based on the order

(i) First-order differential equations: The differential equations with a degree of 1.

(ii) Second-order differential equations: The differential equations with a degree of 2.

Degree of Differential Equations

The degree of a differential equation is the exponent of the highest order derivative when the equation is expressed in polynomial form.

Example 1 (Degree 1):
$ \frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} + \frac{dy}{dx} = xy^2 $

Example 2 (Degree undefined):
$ \frac{d^3y}{dx^3} + \sin \frac{dy}{dx} = 0 $

Example 3 (After rewriting to polynomial form):
$ \left(\frac{d^2y}{dx^2}\right)^{1/2} = \left( y + \left(\frac{dy}{dx}\right)^4 \right)^{1/3} $
Rewritten:
$ \left(\frac{d^2y}{dx^2}\right)^3 = \left( y + \left(\frac{dy}{dx}\right)^4 \right)^2 $

Classification of Differential Equations

The classification of differential equations helps in understanding the type and structure of an equation based on order, degree, and the number of variables involved. This categorization is essential for choosing the appropriate method to solve the differential equation efficiently.

Ordinary Differential Equations (ODEs)

An ordinary differential equation is an equation that involves a function $y = y(x)$ of a single independent variable $x$ and its derivatives. Its general form is:

$ \frac{d^n y}{dx^n} + P_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}} + \dots + P_1(x)\frac{dy}{dx} + P_0(x)y = Q(x) $ where $P_0(x), P_1(x), \dots, P_{n-1}(x)$ and $Q(x)$ are functions of $x$ only.

In simple words, an ODE contains derivatives with respect to a single independent variable. They are widely used in physics and engineering.

Examples:
$ \frac{dy}{dx} + xy = \sin x $
$ \frac{d^3y}{dx^3} + 2\frac{dy}{dx} + y = e^x $

Partial Differential Equations (PDEs)

A partial differential equation involves an unknown function $u = u(x_1, x_2, \dots, x_n)$ of two or more independent variables and its partial derivatives. Its general form can be written as: $ F\left(x_1, x_2, \dots, x_n, u, \frac{\partial u}{\partial x_1}, \dots, \frac{\partial^n u}{\partial x_1^{k_1} \partial x_2^{k_2} \dots \partial x_n^{k_n}} \right) = 0 $ where $k_1 + k_2 + \dots + k_n \le n$ denotes the order of the PDE.

Examples:
$ \frac{\partial u}{\partial y} = -\frac{\partial u}{\partial x} $
$ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 $
$ \frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial t^2} - 2 \frac{\partial u}{\partial t} $

Homogeneous Differential Equations

Any differential equation of the form $M(x, y) d x+N(x, y) d y=0$ or $\frac{d y}{d x}=-\frac{M(x, y)}{N(x, y)}$ is called homogeneous differential equations if $M(x, y)$ and $N(x, y)$ are homogeneous functions of the same degree.

Since, $M(x, y)$ and $N(x, y)$ are both homogeneous function of degree $n$, then $\mathrm{DE}$ can be reduced to a function of $\mathrm{y} / \mathrm{x}$
$\frac{d y}{d x}=-\frac{M(x, y)}{N(x, y)}=\phi\left(\frac{y}{x}\right)$

Non-Homogeneous Differential Equations

Any differential equation which is not homogeneous is called a Non-Homogeneous Differential Equation. A non-homogeneous differential equation of second order has the form: $y^{\prime \prime}+a(t) y^{\prime}+b(t) y=c(t)$

Here, $y^{\prime \prime}$ denotes the second derivative of $y$, and $c(t)$ is a non-zero function of $t$. This equation can be converted to a homogeneous differential equation, and the related DE is $y^{\prime \prime}+a(t) y^{\prime}+b(t) y=0$.

This equation is also called the complementary equation to the given non-homogeneous differential equation.

Linear Differential Equations

The linear differential equations is a linear equation that involves one or more terms consisting of derivatives of the dependent variable concerning one or more independent variables.

The general equation of the first-order differential equations is the form of

$
\frac{d y}{d x}+P(x) \cdot y=Q(x)
$
Where $P(x)$ and $ Q(x)$ are functions of $x$ only or constants.

Non-Linear Differential Equations

A non-linear differential equation is a differential equation in which the unknown function or its derivatives appear non-linearly. This means the dependent variable $y$ and its derivatives are not raised only to the first power, nor do they appear only in linear combinations.

Mathematical definition: A differential equation of the form

$ F\left(x, y, y', y'', \dots, y^{(n)}\right) = 0 $

is non-linear if $F$ is a non-linear function of $y, y', \dots, y^{(n)}$.

Examples: $ \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + y = 0 $

$ y'' + \sin(y) = 0 $

These equations cannot generally be solved by methods used for linear differential equations and often require special techniques or numerical methods.

Exact Differential Equations

The equation $\mathrm{A}(\mathrm{x}, \mathrm{y}) \mathrm{dx}+\mathrm{B}(\mathrm{x}, \mathrm{y}) \mathrm{dy}=0$ is an exact differential equation if there exists a function of two variables $x$ and $y$ having continuous partial derivatives such that the exact differential equation definition is separated as follows: $u_x(x, y)=A(x, y) \text { and } u_y(x, y)=B(x, y)$

General Form: The general form of the exact differential equation is $A(x, y) d x+B(x, y) d y=0$ where $A$ and $B$ are the polynomial functions in terms of $x$ and $y$.

If the differential equation $A(x, y) d x+B(x, y) d y=0$ is not exact, it is possible to make it exact by multiplying using a relevant factor $\mathrm{u}(\mathrm{x}, \mathrm{y})$ which is known as integrating factor for the given differential equation.

Consider an example, $2 y d x+x d y=0$

Now check it whether the given differential equation is exact using testing for exactness.
The given differential equation is not exact.
In order to convert it into the exact differential equation, multiply by the integrating factor $u(x, y)=x$, the differential equation becomes,

$
2 x y d x+x^2 d y=0
$

The above resultant equation is an exact differential equation because the left side of the equation is a total differential of $x^2 y$.

Sometimes it is difficult to find the integrating factor. But, there are two classes of differential equations whose integrating factor may be found easily. Those equations have the integrating factor having the functions of either $x$ alone or $y$ alone.

When you consider the differential equation $A(x, y) d x+B(x, y) d y=0$, the two cases involved are:

Case 1: If $[1 / B(x, y)]\left[A_y(x, y)-B_x(x, y)\right]=h(x)$, which is a function of $x$ alone, then $e^{\int h(x) d x}$ is an integrating factor.

Case 2: If $[1 / A(x, y)]\left[B_x(x, y)-A_y(x, y)\right]=k(y)$, which is a function of $y$ alone, then $e^{j k(y) d y}$ is an integrating factor.

How to Solve Differential Equations?

Solving differential equations can be classified according to different types of equations and given conditions. There are different types of solutions for differential equations based on their types or structures of differential equations. General solutions and particular solutions are two types of differential equation solutions.

A general solution of a differential equation is a relation between the variables (not involving the derivatives) which contains the same number of the arbitrary constants as the order of the differential equation.

A particular solution of the differential equation is obtained from the general solution by assigning particular values to the arbitrary constant in the general solution.

A differential equation solution is a relation between the variables of the equation that satisfy the D.E. and does not contain any derivatives or any arbitrary constants. Now, let us see how to solve differential equations.

To solve the first-order differential equation of first degree, some standard forms are available to get the general solution. They are:

• Variable separable method
• Reducible into the variable separable method by substitution
• Linear differential equations
• Homogenous differential equations
• Exact differential equations

Variable Separable Method

Differential equations where the variables can be separated from each other are called separable differential equations. The general form of a separable differential equation is $\frac{dy}{dx} = f(x)g(y)$.

These equations can be easily solved by separating the variables and integrating them individually.

The differential of the form $\frac{d y}{d x}=f(x) g(y)$ where $f(x)$ is a function of $x$ and $g(y)$ is a function of $y$, are said to be variable separable form.

Steps to Solve differential equations using variable separable methods:

1. Check for any values of y that make $g(y)=0$. These correspond to constant solutions.
2. Rewrite the differential equation in the form $\frac{d y}{g(y)}=f(x) d x$
3. Integrate both sides of the equation.
4. Solve the resulting equation for $y$ if possible.
5. If an initial condition exists, substitute the appropriate values for $x$ and $y$ into the equation and solve for the constant.

Rewrite the equation as
$\frac{d y}{g(y)}=f(x) d x \quad[\text { where } g(y) \neq 0]$

This process is separating the variables. Now, integrating both sides, we get
$
\int \frac{d y}{g(y)}=\int f(x) d x+c
$

By this, we get the solution of the differential equation

Let's see some illustration for a better understanding.
Solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\mathrm{e}^{\mathrm{x}}+1\right)\left(\mathrm{y}^2+1\right)$

Rewrite the differential equation as
$
\frac{\mathrm{dy}}{1+\mathrm{y}^2}=\left(\mathrm{e}^{\mathrm{x}}+1\right) \mathrm{dx}
$

Integrating both sides, we get
$
\begin{aligned}
& \int \frac{\mathrm{dy}}{1+\mathrm{y}^2}=\int\left(\mathrm{e}^{\mathrm{x}}+1\right) \mathrm{dx} \\
& \Rightarrow \tan ^{-1} y=e^x+x+c \\
& \Rightarrow y=\tan \left(e^x+x+c\right)
\end{aligned}
$

Reducible into Variable Separable Method using Substitution

A differential equation of the form $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}(\mathrm{ax}+\mathrm{by}+\mathrm{c})$ where $\mathrm{a}, \mathrm{b}$, and $\mathrm{c}$ are constants, can be converted into an equation with variables separable by the substitution $\mathrm{v}=\mathrm{ax}+\mathrm{by}+\mathrm{c}$.
$
\begin{aligned}
& \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}(\mathrm{ax}+\mathrm{by}+\mathrm{c}) \\
& \mathrm{v}=\mathrm{ax}+\mathrm{by}+\mathrm{c} \\
& \therefore \frac{d v}{d x}=a+b \frac{d y}{d x} \text { or, } \frac{d y}{d x}=\frac{\frac{d v}{d x}-a}{b} \\
& \Rightarrow \frac{\frac{\mathrm{d} v}{\mathrm{dx}}-\mathrm{a}}{\mathrm{b}}=\mathrm{f}(\mathrm{v}) \Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{bf}(\mathrm{v})+\mathrm{a} \\
& \Rightarrow \frac{\mathrm{dv}}{\mathrm{bf}(\mathrm{v})+\mathrm{a}}=\mathrm{dx} \quad ...(ii)
\end{aligned}
$

In the differential equation (ii), the variables $\mathrm{x}$ and $\mathrm{v}$ are separated.
Integrating (ii), we get
$
\begin{aligned}
& \Rightarrow \quad \int \frac{d v}{b f(v)+a}=\int d x+C \\
& \Rightarrow \quad \int \frac{d v}{b f(v)+a}=x+C \text {, where } v=a x+b y+c
\end{aligned}
$

This represents the general solution of the differential equation (i).

Example: Solve $y^{\prime}=\sin ^2(x-y+1)$.
Put $z=x-y+1$, so that $\frac{d z}{d x}=1-\frac{d y}{d x}$.
Thus, the given equation reduces to $1-\frac{d z}{d x}=\sin ^2 z$.
i.e., $\frac{d z}{d x}=1-\sin ^2 z=\cos ^2 z$.

Separating the variables leads to $\frac{d z}{\cos ^2 z}=d x$ (or) $\sec ^2 z d z=d x$.
On integration, we get $\tan z=x+C$ (or) $\tan (x-y+1)=x+C$.

Linear Differential Equations

To solve these types of equations, various methods can be used:

1. Separation of Variables: Applicable to those types of equations which is very simple in which variables can be separated easily on each side of the equation.
2. Integrating Factor: For first-order(degree) linear equations of the form dy/dx+P(x)y=Q(x), we multiply by an integrating factor to simplify.
3. Characteristic Equation: For constant coefficient linear differential equations, we solve the characteristic polynomial to find the solutions of the equations.

Solve Linear Differential Equation using Integrating Factor

Integrating factor: A term, which when multiplied by an expression, converts it to an exact differential i.e. a function which is the derivative of another function.

We have, $\frac{d y}{d x}+P(x) \cdot y=Q(x)$
multiply both sides of Eq (i) by $\int e^{P(x) d x}$, we get
i.e. $\quad e^{\int P(x) d x} \cdot \frac{d y}{d x}+y \cdot P(x) \frac{d}{d x}\left(e^{\int P(x) d x}\right)=Q e^{\int P(x) d x}$
or $\quad \frac{d}{d x}\left(\mathrm{ye}^{\int P(x) d x}\right)=\mathrm{e}^{\int P(x) d x} \cdot Q(x)$
Integrating both sides, we get
or $\quad \int \mathrm{d}\left(y e^{\int P(x) d x}\right)=\int\left(e^{\int P(x) d x} \cdot Q(x)\right) d x$
$\Rightarrow \quad y \mathrm{e}^{\int P(x) d x}=\int \mathrm{Q}(\mathrm{x}) \mathrm{e}^{\int \mathrm{P}(\mathrm{x}) d \mathrm{dx}} \mathrm{dx}+\mathrm{C}$

Which is the required solution of the given differential equation.

The term $\mathrm{e}^{/ \mathrm{P}(\mathrm{x}) \mathrm{dx}}$ which convert the left hand expression of the equatio into a perfect differential is called an Integrating factor (IF).

Thus, we remember the solution of the above equation as

$
y(\mathrm{IF})=\int Q(\mathrm{IF}) d x+C
$

Note: Sometimes a given differential equation can be made linear if we take $x$ as the dependent variable and $y$ as the independent variable. So, we can check the equation with respect to both $x$ and $y$.

Example: Solve $\frac{d y}{d x}+2 y \cot x=3 x^2 \operatorname{cosec}^2 x$.

Here, $P=2 \cot x ; Q=3 x^2 \operatorname{cosec}^2 x$.

$
\int P d x=\int 2 \cot x d x=2 \log |\sin x|=\log |\sin x|^2=\log \sin ^2 x
$

Thus, I.F $=e^{\int P d x}=e^{\log \sin ^2 x}=\sin ^2 x$.
Hence, the solution is. $y e^{\int_{P d x}}=\int Q e^{\int_{p d x}} d x+C$.
That is, $y \sin ^2 x=\int 3 x^2 \operatorname{cosec}^2 x \cdot \sin ^2 x d x+C=\int 3 x^2 d x+C=x^3+C$.
Hence, $y \sin ^2 x=x^3+C$ is the required solution.

Homogeneous Differential Equations

This equation can be solved by the substitution $\mathrm{y}=\mathrm{vx}$.

$\begin{aligned}
& y = v x \\
& \Rightarrow \quad \frac{\mathrm{d}y}{\mathrm{dx}} = v + x \frac{\mathrm{d}v}{\mathrm{dx}}
\end{aligned}$

Thus, $\frac{d y}{d x}=\phi\left(\frac{y}{x}\right)$ transforms to
$
\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\phi(\mathrm{v})
$

$
\Rightarrow \quad \frac{d v}{\phi(v)-v}=\frac{d x}{x}
$

The variables have now been separated and the solution is
$
\int \frac{\mathrm{dv}}{\phi(\mathrm{v})-\mathrm{v}}=\ln \mathrm{x}+\mathrm{c}
$

After the integration $\mathrm{v}$ should be replaced by $\mathrm{y} / \mathrm{x}$ to get the required solution.

If the differential equation is of the form
$
\frac{d y}{d x}=\frac{a x+b y+c}{d x+e y+f} \quad ...(1)
$

It can be reduced to a homogeneous differential equation as follows:
Put $x=X+h, y=Y+k \quad ...(2)$
where $\mathrm{X}$ and $\mathrm{Y}$ are new variables and $\mathrm{h}$ and $\mathrm{k}$ are constants yet to be chosen
From (2)
$
d x=d X, d y=d Y
$

Equation (1), thus reduces to
$
\frac{d Y}{d X}=\frac{a(X+h)+b(Y+k)+c}{d(X+h)+e(Y+k)+f}=\frac{a X+b Y+(a h+b k+c)}{d X+e Y+(d h+e k+f)} \quad ...(3)
$

In order to have equation (3) as a homogeneous differential equation, choose $\mathrm{h}$ and $\mathrm{k}$ such that the following equations are satisfied :
$
\left.\begin{array}{rl}
a h+b k+c & =0 \\
d h+e k+f & =0
\end{array}\right\}
$

Now, (3) becomes
$
\frac{d Y}{d X}=\frac{a X+b Y}{d X+e Y}
$
which is a homogeneous differential equation and can be solved by putting $Y=v X$.

Separate the variables and integrate them to get the required solution.

Example: Solve $\left(x^2-3 y^2\right) d x+2 x y d y=0$.

Now, we rewrite the given equation as $\frac{d y}{d x}=\frac{3 y}{2 x}-\frac{x}{2 y}$.
Taking $y=v x$, we have $v+x \frac{d v}{d x}=\frac{3 v}{2}-\frac{1}{2 v}$ or $x \frac{d v}{d x}=\frac{v^2-1}{2 v}$.Separating the variables, we obtain $\frac{2 v d v}{v^2-1}=\frac{d x}{x}$.

On integration, we get $\log \left|v^2-1\right|=\log |x|+\log |C|$,
Hence $\left|v^2-1\right|=|C x|$, where $C$ is an arbitrary constant.
Now, replace $v$ by $\frac{y}{x}$ to get $\left|\frac{y^2}{x^2}-1\right|=|C x|$.
Thus, we have $\left|y^2-x^2\right|=\left|C x^3\right|$.
Hence, $y^2-x^2= \pm C x^3$ (or) $y^2-x^2=k x^3$ gives the general solution.

Exact Differential Equations

The following steps explain how to solve the exact differential equation:

Step 1: The first step to solving the exact differential equation is to make sure the given differential equation is exact.

$
\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}
$

Step 2: Write the system of two differential equations that defines the function $u(x, y)$. That is

$
\begin{aligned}
& \frac{\partial u}{\partial x}=P(x, y) \\
& \frac{\partial u}{\partial y}=Q(x, y)
\end{aligned}
$

Step 3: Integrating the first equation over the variable $x$, we get

$
u(x, y)=\int P(x, y) d x+\phi(y)
$

Step 4: Differentiating concerning $y$, substitute the function $u(x, y)$ in the second equation

$
\frac{\partial u}{\partial x}=\frac{\partial}{\partial x}\left[\int P(x, y) d x+\phi(y)\right]=Q(x, y)
$
Step 5: We can find the function $\varphi(y)$ by integrating the last expression so that the function $\mathrm{u}(\mathrm{x}, \mathrm{y})$ becomes

$
u(x, y)=\int P(x, y) d x+\phi(y)
$

Step 6: Finally, the general solution of the exact differential equation is given by

$
u(x, y)=C
$

Illustration 1: Solution of the differential equation $2 x y d x+\left(x^2+3 y^2\right) d y=0$ is Let us first separate terms containing only $x$ with $dx$ and terms containing only $y$ with $dy$

$
2 x y d x+x^2 d y+3 y^2 d y=0
$

Here first two terms have both $x$ and $y$. We can make an observation that first two terms are the differentiation of $x^2 y$. Hence we can write this equation as

$
d\left(x^2 y\right)+3 y^2 d y=0
$

Integrating this, we get

$
x^2 y+y^3+c=0
$
This is the solution of this equation

Illustration 2: $\frac{x d y-y d x}{x^2+y^2}+e^x d x=0$

Observe that

$
\frac{x d y-y d x}{x^2+y^2}=\frac{\frac{x d y-y d x}{x^2}}{1+\frac{y^2}{x^2}}=\frac{d\left(\frac{y}{x}\right)}{1+\left(\frac{y}{x}\right)^2}=d\left[\tan ^{-1}\left(\frac{y}{x}\right)\right]
$

So the equation is

$
d\left[\tan ^{-1} \frac{y}{x}\right]+e^x d x=0
$
Integrating

$\tan ^{-1} \frac{y}{x}+e^x+c=0$

Important Books and Resources for Differential Equations

This section highlights the best books and reference materials to strengthen your understanding of class 12 differential equations and practice effectively for exams.

NCERT Resources

Here, you’ll find NCERT resources that explain differential equations concepts clearly and form the foundation for solving problems.

• NCERT Maths Notes for Class 12th Chapter 9 - Differential Equations

• NCERT Maths Solutions for Class 12th Chapter 9 - Differential Equations

• NCERT Maths Exemplar Solutions for Class 12th Chapter 9 - Differential Equations

NCERT Subjectwise Resources

Explore subject-wise NCERT resources that cover notes, solutions and exemplar solutions, supporting structured and focused learning.

Practice Questions based on Differential Equations

This section provides important practice questions and exercises on differential equations, helping you improve problem-solving skills and exam readiness.