Differential Equations - Topics, Types, Books, FAQs

A differential equation is an equation in which some derivatives of the unknown function occur. Many real-life problems concern relationships between changing quantities. Since rates of change are represented mathematically by derivatives, many mathematical models often involve equations relating to an unknown function with one or more of its derivatives. Such equations are differential equations.

They are of basic significance in science and engineering since many physical laws as well as relations are modelled in the form of differential equations. Differential equations are much useful in describing mathematical models involving population growth or radioactive decay. The study of biological sciences and economics is incomplete without the application of differential equations. This article is about the concept of Differential Equations Class 12.

Differential Equations

A differential equation is any equation which contains at least one derivative of an unknown function.

The general form of differential equation is $\frac{d y}{d x}=f(x)$ where " $x$ " is an independent variable and " $y$ " is a dependent variable

Example of differential equations: $x \frac{d x}{d x}+2 y=0$

The above-written equation involves variables as well as the derivative of the dependent variable $y$ with respect to the independent variable $x$. Therefore, it is a differential equation.

The following relations are some of the examples of differential equations:

(i) $\frac{d y}{d x}=\sin 2 x+\cos x$
(ii) $k \frac{d^2 y}{d x^2}=\left[1+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2}$

Order of Differential Equations

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

Some examples of different orders of the differential equations are given:

(i) $\frac{\mathrm{dy}}{\mathrm{dx}}=\sin 2 \mathrm{x}+\cos \mathrm{x}$ is of order $1$
(ii) $\frac{d^2 y}{d x^2}+y=0$ is of order $2$
(iii) $\frac{d^3 y}{d x^3}+2\left(\frac{d^2 y}{d x^2}\right)^5+y=e^x$ is of order $3$

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

If a differential equation is expressed in a polynomial form, then the integral power of the highest order derivative appears is called the degree of the differential equation.

Consider the following example:

$\begin{aligned} & \frac{d^3 y}{d x^3}+2\left(\frac{d^2 y}{d x^2}\right)^2+\frac{d y}{d x}=x y^2 \\ & \frac{d^3 y}{d x^3}+\sin \frac{d y}{d x}=0\end{aligned}$

Observe that equation (i) is a polynomial equation in derivatives: $y''', y'', and y' $ in this case: these terms occur only with some whole number power. So for this equation, the degree of differential equations can be defined.

But equation (ii) is not a polynomial in $y’ $, as $\sin (y')$ term is present, so the degree of this equation is not defined.

Now, consider another example,
$
\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{2}}=\left(y+\left(\frac{d y}{d x}\right)^4\right)^{\frac{1}{3}}
$

Here, derivatives have no integral power. We can rewrite the equation as
$
\left(\frac{d^2 y}{d x^2}\right)^3=\left(y+\left(\frac{d y}{d x}\right)^4\right)^2
$

After the expansion of the above equation, this will become a polynomial equation of derivatives. So, its degree is defined.

So, the degree of (i) is $1$, as the power of the highest order derivative ( $y^{\prime \prime \prime}$ ) is $1$.

And the degree of
$
\left(\frac{d^2 y}{d x^2}\right)^3=\left(y+\left(\frac{d y}{d x}\right)^4\right)^2
$
is $3$ (the power of the highest order derivative, which is $\left.y^{\prime \prime}\right)$.

Classification of Differential Equations

The types of differential equations are

Ordinary Differential Equations

If a differential equation contains only ordinary derivatives of one or more functions with respect to a single independent variable, it is said to be an Ordinary Differential Equation (ODE)
For example
(i) $\frac{\mathrm{dy}}{\mathrm{dx}}+x y=\sin \mathrm{x}$
(ii) $\frac{\mathrm{d}^3 \mathrm{y}}{\mathrm{dx}^3}+2 \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=\mathrm{e}^{\mathrm{x}}$

Partial Differential Equations

The equations involving only partial derivatives of one or more functions of two or more independent variables are called Partial Differential Equations (PDE).

For instance, $\frac{\partial u}{\partial y}=-\frac{\partial u}{\partial x}, \quad \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$ and $\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial t^2}-2 \frac{\partial u}{\partial t}$ are some examples of partial differential equations.

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 Homogenous is called a Non-Homogenous 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

When the equation is not linear in an unknown function and its derivatives, then it is said to be a nonlinear differential equation.

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)
$

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.

Differential Equations Formulas Class 12

Differential equation formulas include the formulas for solving 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.

Differential equations solutions 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 Saparable 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

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

Related Articles

• Formation Of Differential Equation And Solutions Of A Differential Equation

• Application Of Differential Equation

Importance of Class 12 Differential Equations

Differential Equations have a significant weighting in the IIT JEE test, which is a national level exam for 12th grade students that aids in admission to the country's top engineering universities. It is one of the most difficult exams in the country, and it has a significant impact on students' futures. When it comes to math, the significance of these chapters cannot be overstated due to their great weightage. You may begin and continue your studies with the standard books and these revision notes, which will ensure that you do not miss any crucial ideas and can be used to revise before any test or actual examination.

How to Study Differential Equations Class 12?

Start with understanding the basic concepts of Differentiation which play an important role in this chapter. Once you're clear with basic concepts, move to important or standard formulas used in differentiation. After which start practicing to identify the type of differential equation and try solving it. Practice as many differential equations examples and solutions as you can.

If you are preparing for competitive exams then solve as many problems as you can. Do not jump on the solution right away. Remember if your basics are clear you should be able to solve any question on this topic. At the end of the chapter remember to make short notes on the method of solutions of different differential equations to revise quickly before exams or anytime when you are required to revise the chapter.

Best Books for Differential Equations Class 12

Start from NCERT Books, the illustration is simple and lucid. You should be able to understand most of the things. Solve all problems (including miscellaneous problem) of NCERT. If you do this, your basic level of preparation will be completed.

Then you can refer to the book Calculus by Dr. SK goyal or RD Sharma. Differential equations are explained very well in these books and there are an ample amount of questions with crystal clear concepts. Choice of reference book depends on person to person, find the book that best suits you the best, depending on how well you are clear with the concepts and the difficulty of the questions you require.

NCERT Solutions Subject-wise link:

• NCERT solutions for class 12 Physics.
• NCERT solutions for class 12 Chemistry.
• NCERT solutions for class 12 Mathematics.
• NCERT solutions for class 12 Biology.