## What is Differential Equation?

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differential equations.

The weight and height of a normal child will increase as it keeps growing old. How will you say this mathematically? The rate in which the weight of the child increases is dependent to the child’s height. The rate in which something changes is written as a derivative. One quantity is dependent on the other. The relationship between the dependent variable, independent variable and their derivatives is called a differential equation.

## Differential Equations with One Independent Variable (ODE)

The differential equation where the dependent variable depends on only one independent variable is called as an ordinary differential equation (ODE). The notations used for the derivatives are $\frac{dy}{dx},\frac{{d}^{2}y}{d{x}^{2}},{y}^{\prime },{y}^{″}$, etc.

## Differential Equation with Two or More Independent Variables (PDE)

The differential equation where, there are two or more than two independent variable is called as a partial differential equation (PDE). The notations used for partial derivatives are $\frac{\partial x}{\partial y},\frac{{\partial }^{2}x}{\partial {y}^{2}}$, etc.

## Order of a Differential Equation

It is the order of the highest order derivative present in the equation.

For example-

$2{x}^{2}\frac{{d}^{2}y}{d{x}^{2}}+4x\frac{dy}{dx}+y=0$

Here, order of highest order derivative is 2. So, the order is 2.

## Degree of a Differential Equation

It is the power of the highest order derivative in the differential equation.

For example-

${\left(\frac{dy}{dx}\right)}^{2}+\frac{dy}{dx}=4$

Here, power of the derivative is 2. So, the degree is 2.

The degree should be a positive integer.

## How is a Differential Equation Formed?

It can be formed from its general solution. The order depends on the arbitrary constants in the general solution.

For example,

Consider, $y=A{x}^{2}+C$ where, A and C are arbitrary constants.

As there are 2 arbitrary constants, the differential equation has order 2. Differentiate the equation 2 times in x terms.

So,

$\begin{array}{l}y=A{x}^{2}+C\\ \frac{dy}{dx}=2Ax\end{array}$

Again differentiate in x terms.

$\frac{{d}^{2}y}{d{x}^{2}}=2A$

Multiply the equation by x

$x\frac{{d}^{2}y}{d{x}^{2}}=2Ax$

Substitute $\frac{dy}{dx}=2Ax$

$x\frac{{d}^{2}y}{d{x}^{2}}=\frac{dy}{dx}$

Hence the differential equation is $x\frac{{d}^{2}y}{d{x}^{2}}=\frac{dy}{dx}$

## Solution of the Ordinary Differential Equations

The solution is the equation which does not contain derivative of variables which are dependent or independent. It satisfies the differential equation.

The two types of solution for a differential equation are as follows.

1) General solution

2) Particular solution

In general solution, the number of arbitrary constants and the order of differential equation are equal.

The particular solution is obtained by putting values of arbitrary constant in the general solution.

Let us take a differential equation $\frac{dy}{dx}=\mathrm{cos}x$.

Here, the solution is $y=\mathrm{sin}x+c$ and c is an arbitrary constant. So the differential equation has infinitely many solutions.

$y=\mathrm{sin}x+c$is the general solution and by replacing value of $c=3$ in general solution $y=\mathrm{sin}x+3$ it becomes a particular solution.

### Variable Separation Method to solve differential equation

Variable separation is to rearrange the terms of equation as, all one variable dependent term in a side and another variable dependent term on other side and to solve it.

Suppose the first order and the first degree differential equation is written in the form of $Adx=Bdy$

Where, A is function of x and B is function of y then this is called as variable separable form.

The solution can be found by integrating both sides.

$\int Adx+\int Bdx=c$

Here, c is an arbitrary constant.

### Homogenous function

If $f\left(\lambda x,\lambda y\right)={\lambda }^{n}f\left(x,y\right)$where, $\lambda$is a constant then the function $f\left(x,y\right)$ is called as a homogeneous function.

### Homogenous differential equation

If $f\left(x,y\right)$and $g\left(x,y\right)$are homogenous functions of same degree then $f\left(x,y\right)dx+g\left(x,y\right)dy=0$is called as homogenous differential equation.

Method of solving is given below.

$f\left(x,y\right)dx+g\left(x,y\right)dy=0$ is a homogenous differential equation.

First, express the equation in form:

$\frac{dy}{dx}=\frac{-f\left(x,y\right)}{g\left(x,y\right)}$

Substitute $y=ux$:

$\frac{dy}{dx}=u+x\frac{du}{dx}$

Now, the solution is in two variables x and u.

Substitute back $u=\frac{y}{x}$

For example,

${x}^{2}ydx-\left({x}^{3}+{y}^{3}\right)dy=0$ …… (1)

Simplify the equation:

$\begin{array}{l}{x}^{2}ydx=\left({x}^{3}+{y}^{3}\right)dy\\ \frac{dy}{dx}=\frac{{x}^{2}y}{{x}^{3}+{y}^{3}}\end{array}$

Put $y=ux$.

Differentiate y in x terms:

$\frac{dy}{dx}=u+x\frac{du}{dx}$ …… (2)

Substitute $y=ux$ and equation (2) in equation (1):

$\begin{array}{c}u+x\frac{du}{dx}=\frac{{x}^{2}\cdot ux}{{x}^{3}+{u}^{3}{x}^{3}}\\ =\frac{{x}^{3}\cdot u}{{x}^{3}\left(1+{u}^{3}\right)}\\ =\frac{u}{\left(1+{u}^{3}\right)}\end{array}$

$\begin{array}{l}x\frac{du}{dx}=\frac{u}{1+{u}^{3}}-u\\ x\frac{du}{dx}=\frac{-{u}^{4}}{1+{u}^{3}}\end{array}$

Separate the variable:

$\frac{1+{u}^{3}}{{u}^{4}}du=-\frac{dx}{x}$

Integrate on both sides:

$\begin{array}{l}\int \frac{1+{u}^{3}}{{u}^{4}}du=-\int \frac{dx}{x}\\ \int \left(\frac{1}{{u}^{4}}+\frac{1}{u}\right)du=-\mathrm{log}\left|x\right|+c\\ -\frac{1}{3{u}^{3}}+\mathrm{log}\left|u\right|=-\mathrm{log}\left|x\right|+c\end{array}$

Substitute back $u=\frac{y}{x}$:

$\begin{array}{l}-\frac{1}{3{\left(\frac{y}{x}\right)}^{3}}+\mathrm{log}\left|\frac{y}{x}\right|=-\mathrm{log}\left|x\right|+c\\ -\frac{{x}^{3}}{3{y}^{3}}+\mathrm{log}\left|y\right|-\mathrm{log}\left|x\right|=-\mathrm{log}\left|x\right|+c\\ \mathrm{log}\left|y\right|-\frac{{x}^{3}}{3{y}^{3}}=c\end{array}$

### Linear Differential equation

If the variable which is dependent and its differential coefficient are of degree 1, then it is a linear differential equation.

$\frac{dy}{dx}+Ay=B$

Here, A and B are functions of x.

Method to solve first order linear equation

First, find integrating factor (IF) ${e}^{\int Adx}$.

Write solution in form:

$y\left({e}^{\int Adx}\right)=\int \left(B×{e}^{\int Adx}\right)dx+c$

For example:

$\frac{dy}{dx}+y={e}^{-x}$

Compare the equation with $\frac{dy}{dx}+Ay=B$:

So $A=1$and $B={e}^{-x}$.

Find IF:

$\begin{array}{c}IF={e}^{\int Adx}\\ ={e}^{\int dx}\\ ={e}^{x}\end{array}$

Write the solution:

$\begin{array}{l}y\left({e}^{x}\right)=\int {e}^{-x}{e}^{x}dx+c\\ y{e}^{x}=x+c\\ y=\frac{\left(x+c\right)}{{e}^{x}}\end{array}$

## Applications of Differential Equation

Differential equation has wide range of applications such as rate change problems, physical problems, and electrical problems.

### Population growth and growth of bacteria

The population of a country increase with time.

Suppose P is population of the country at time t. Then the rate of increase in population of the country is proportional to original population.

$\begin{array}{l}\frac{dP}{dt}\propto P\\ \frac{dP}{dt}=kP\end{array}$

Where, k is proportionality constant.

By integrating it gives population at time t.

The rate of disintegration of radioactive element is always proportional to the amount present at time t.

If z is the amount of radioactive material present at time t, then,$\begin{array}{l}\frac{dz}{dt}\propto -z\\ \frac{dz}{dt}=-kz\end{array}$

Where, k is proportionality constant. Negative sign represents that x decreases as the time increases.

After solving differential equation it gives the amount of radioactive substance at time t.

Half-life period - Time taken for half of substance to disintegrate is called as half-life period.

Example:

The rate of the disintegration of the radioactive element at time t is proportional to its mass. Find the time during which the original mass of $1.5gm$will disintegrate into its mass of $0.5gm$.

Let x be the mass of the radioactive element.

$\frac{dx}{dt}=-kx$

Separate the variable:

$\frac{dx}{x}=-kdt$

Integrate on both sides:

$\begin{array}{l}\int \frac{dx}{x}=\int -kdt\\ \mathrm{log}x=-kt+c\end{array}$ …… (1)

Substitute $t=0,\text{\hspace{0.17em}}x=1.5$

$\begin{array}{l}\mathrm{log}1.5=-k\left(0\right)+c\\ \mathrm{log}1.5=c\end{array}$

Substitute $x=0.5$and $\mathrm{log}1.5=c$in equation (1)

$\begin{array}{l}\mathrm{log}\left(0.5\right)=-kt+\mathrm{log}\left(1.5\right)\\ kt=\mathrm{log}\left(1.5\right)-\mathrm{log}\left(0.5\right)\\ kt=\mathrm{log}\left(\frac{1.5}{0.5}\right)\\ t=\frac{\mathrm{log}3}{k}\end{array}$

The original mass of $1.5gm$will disintegrate into its mass of $0.5gm$in $\frac{\mathrm{log}3}{k}$ units of time.

### Newton’s law of cooling

The rate of change of temperature of heated body at any time is proportional to the difference between temperature of the body and surrounding medium.

Suppose $\theta$ is the temperature of the body at time t and $\phi$be the temperature of the surrounding.

$\begin{array}{l}\frac{d\theta }{dt}\propto \left(\theta -\phi \right)\\ \frac{d\theta }{dt}=-k\left(\theta -\phi \right)\end{array}$

Where, k is proportionality constant. Negative sign is there as the temperature decreases.

After solving differential equation it gives temperature of the body at time t.

Example:

If a body cools from ${80}^{\circ }C$to ${50}^{\circ }C$ at room temperature of ${25}^{\circ }C$in 30 minutes, Find temperature of the body after 1 hour.

Let $\theta$ be the temperature of the body at time t.

So,

$\frac{d\theta }{dt}=-k\left(\theta -25\right)$

Separate the variables:

$\frac{d\theta }{\theta -25}=kdt$

Integrate on both sides:

$\begin{array}{l}\int \frac{d\theta }{\theta -25}=\int kdt\\ \mathrm{log}\left(\theta -25\right)=kt+c\end{array}$ …… (1)

Substitute $t=0,\text{\hspace{0.17em}}\theta ={80}^{\circ }$:

$\begin{array}{l}\mathrm{log}\left(\theta -25\right)=k\left(0\right)+c\\ \mathrm{log}\left(80-25\right)=c\\ \mathrm{log}55=c\end{array}$

Substitute $t=30,\text{\hspace{0.17em}}\theta =50$and $\mathrm{log}55=c$in equation (1):

$\begin{array}{l}\mathrm{log}\left(50-25\right)=30k+\mathrm{log}\left(55\right)\\ \mathrm{log}\left(25\right)-\mathrm{log}\left(55\right)=30k\\ \mathrm{log}\left(\frac{25}{55}\right)=30k\\ k=\frac{1}{30}\mathrm{log}\left(\frac{5}{11}\right)\end{array}$

Substitute $t=60$, then,

$\mathrm{log}\left(\theta -25\right)=60k+\mathrm{log}\left(55\right)$

Substitute $k=\frac{1}{30}\mathrm{log}\left(\frac{5}{11}\right)$

$\begin{array}{l}\mathrm{log}\left(\theta -25\right)=60\left(\frac{1}{30}\mathrm{log}\left(\frac{5}{11}\right)\right)+\mathrm{log}\left(55\right)\\ =2\mathrm{log}\left(\frac{5}{11}\right)+\mathrm{log}\left(55\right)\\ =\mathrm{log}\frac{125}{11}\end{array}$

Find $\theta$:

$\begin{array}{l}\theta -25=\frac{125}{11}\\ \theta ={36.36}^{\circ }C\end{array}$

Temperature of the body after 1 hour is ${36.36}^{\circ }C$.

## Formulas

• Power rule for differentiation: $\frac{d}{dx}\left({x}^{m}\right)=m{x}^{m-1}$, where $m$ is a real number.
• Product rule for differentiation: $\frac{d}{dx}\left(uv\right)=v\frac{d}{dx}\left(u\right)+u\frac{d}{dx}\left(v\right)$, where $u$ and $v$ are differentiable functions in $x$.
• Power rule for integration: , where $m$ is a real number.

## Practice Problems

1. Find the order of the differential equation:

$4x\frac{{d}^{2}y}{d{x}^{2}}+\frac{dy}{dx}+y=0$

Solution:

The order of highest order derivative is 2. So, the order is 2.

2. Find the degree of the differential equation:

$4x{\left(\frac{{d}^{2}y}{d{x}^{2}}\right)}^{3}+\frac{dy}{dx}+y=0$

Solution:

The power of the highest order derivative is 3. So, the degree is 3.

3. Find the differential equation for $y=P{e}^{3x}+Q{e}^{-3x}$.

Solution:

The given equation has 2 arbitrary constants $P$ and $Q$. Differentiate the equation 2 times with respect to $x$.

So,

$\begin{array}{c}\frac{dy}{dx}=3P{e}^{3x}-3Q{e}^{-3x}\\ =3\left(P{e}^{3x}-Q{e}^{-3x}\right)\end{array}$

Again differentiate with respect to $x$.

$\begin{array}{c}\frac{{d}^{2}y}{d{x}^{2}}=3\left(3P{e}^{3x}+3Q{e}^{-3x}\right)\\ =9\left(P{e}^{3x}+Q{e}^{-3x}\right)\end{array}$

Substitute $y$ for $P{e}^{3x}+Q{e}^{-3x}$ in the second derivative.

$\frac{{d}^{2}y}{d{x}^{2}}=9y$

Hence, the differential equation is $\frac{{d}^{2}y}{d{x}^{2}}=9y$.

4. Find the solution of $\frac{{d}^{2}y}{d{x}^{2}}+x=0$.

Solution:

Simplify the given equation:

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

Integrate on both sides of the equation:

$\begin{array}{l}\int \frac{{d}^{2}y}{d{x}^{2}}=-\int x\\ \frac{dy}{dx}=-\frac{{x}^{2}}{2}+{c}_{1}\end{array}$

Again integrate on both sides:

$\begin{array}{l}\int \frac{dy}{dx}=-\int \frac{{x}^{2}}{2}+\int {c}_{1}\\ y=-\frac{{x}^{3}}{6}+{c}_{1}x+{c}_{2}\end{array}$

where, ${c}_{1}$ and ${c}_{2}$ are arbitrary constants.

5. Solve ${y}^{2}dx+\left(xy+{x}^{2}\right)dy=0$.

Solution:

Simplify the equation:

$\begin{array}{l}{y}^{2}dx=-\left(xy+{x}^{2}\right)dy\\ \frac{dy}{dx}=-\frac{{y}^{2}}{xy+{x}^{2}}\end{array}$ ..….(1)

Put $y=ux$, then differentiate with respect to $x$.$\frac{dy}{dx}=u+x\frac{du}{dx}$ .…..(2)

Substitute $y=ux$and equation 2 in equation 1, then simplify.

$\begin{array}{c}u+x\frac{du}{dx}=-\frac{{u}^{2}{x}^{2}}{x\cdot ux+{x}^{2}}\\ =-\frac{{u}^{2}{x}^{2}}{{x}^{2}\left(u+1\right)}\\ =-\frac{{u}^{2}}{u+1}\\ x\frac{du}{dx}=-\frac{{u}^{2}}{u+1}-u\\ x\frac{du}{dx}=\frac{-2{u}^{2}-u}{u+1}\end{array}$

Separate the variables:

$\frac{u+1}{-2{u}^{2}-u}du=\frac{dx}{x}$

Integrate on both sides:

$\begin{array}{l}\int \frac{u+1}{-2{u}^{2}-u}du=\int \frac{dx}{x}\\ 2\left(-\frac{1}{8}\mathrm{ln}\left|16{u}^{2}+8u\right|+\frac{3}{8}\left(\mathrm{ln}\left|4u+2\right|-\mathrm{ln}\left(4\left|u\right|\right)\right)\right)=\mathrm{ln}\left|x\right|+C\end{array}$

Substitute back $u=\frac{y}{x}$

$\begin{array}{l}2\left(-\frac{1}{8}\mathrm{ln}\left|{\frac{16y}{{x}^{2}}}^{2}+\frac{8y}{x}\right|+\frac{3}{8}\left(\mathrm{ln}\left|\frac{4y}{x}+2\right|-\mathrm{ln}\left(4\left|\frac{y}{x}\right|\right)\right)\right)=\mathrm{ln}\left|x\right|+C\\ -\frac{1}{4}\mathrm{ln}\left|{\frac{16y}{{x}^{2}}}^{2}+\frac{8y}{x}\right|+\frac{3}{4}\left(\mathrm{ln}\left|\frac{4y}{x}+2\right|-\mathrm{ln}\left(8\left|\frac{y}{x}\right|\right)\right)=\mathrm{ln}\left|x\right|+C\end{array}$

6. Solve the differential equation $\left(x+y\right)\frac{dy}{dx}=1$.

Solution:

Obtain the given equation in the form $\frac{dx}{dy}+Ax=B$.

$\begin{array}{l}\frac{dy}{dx}=\frac{1}{x+y}\\ \frac{dx}{dy}=x+y\\ \frac{dx}{dy}-x=y\end{array}$

Compare the simplified equation with the equation $\frac{dx}{dy}+Ax=B$.

$A=-1,B=y$

Find the integrating factor, IF:

$\begin{array}{l}IF={e}^{\int Ady}\\ ={e}^{\int \left(-1\right)dy}\\ ={e}^{-y}\end{array}$

Write the solution:

$\begin{array}{l}x{e}^{-y}=\int y{e}^{-y}dy+C\\ x{e}^{-y}=-y{e}^{-y}-{e}^{-y}+C\\ x=-y-1+C{e}^{y}\\ x+y+1=C{e}^{y}\end{array}$

## Context and Applications

This topic is significant in the professional exams for both undergraduate and graduate courses, especially for

• Bachelor of Science in Mathematics
• Master of Science in Mathematics

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