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Introduction to Differential Equations

An equation which involves derivatives of a dependent variable with respect to another independent variable is called a differential equation. Differential equations are not only used in the field of Mathematics but also play a major role in other fields, such as Medicine, Chemistry, Physics and Engineering. It is a tool which helps in building various mathematical models. In this section, we will study differential equations in detail, along with solved examples.

What Are Differential Equations?

An equation involving a function and one or more of its derivatives is called a differential equation. Differential equations are also defined as the equation that contains derivatives of one or more dependent variables with respect to one or more independent variables. If a function has only one independent variable, then it is an ordinary differential equation.

Examples of Differential Equations:

  1. \(\begin{array}{l}\frac{{{d}^{2}}y}{d{{x}^{2}}}-\frac{dy}{dx}-4x=0\end{array} \)
  2. \(\begin{array}{l}\left( \frac{d{{y}^{2}}}{d{{x}^{2}}} \right)-6\frac{dy}{dx}+8y=0\end{array} \)
  3. \(\begin{array}{l}\left[ 2+\left( \frac{dy}{dx} \right) \right]={{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{\frac{2}{3}}}\end{array} \)

Note: Differential equation involving a function of several variables of its partial derivatives is called a partial differential equation.

Order and Degree of Differential Equations

  • Order: Highest differential coefficient
  • Degree: Differential equation, when expressed as [Polynomial in the derivatives], is the power of the highest order derivative
\(\begin{array}{l}f\left( xy \right)\frac{{{d}^{m}}y}{d{{x}^{m}}}+g\left( xm \right)\frac{{{d}^{m-1}}y}{d{{x}^{m-1}}}+…=0\end{array} \)


The degree of a differential equation is only defined when it is in polynomial form.

The degree may not be defined in ways as mentioned below:

  • \(\begin{array}{l}y’+6{{y}^{2}}+y=0\ \text{at}\ (1, 1)\end{array} \)
  • \(\begin{array}{l}{{\left( y’ \right)}^{2}}+{{y}^{2}}-1=0\ \text{at}\ (1, 2)\end{array} \)
  • \(\begin{array}{l}y”+{{\left( 2{{y}^{1}} \right)}^{6}}+\sin y=0\ \text{at}\ (2, 1)\end{array} \)
  • \(\begin{array}{l}y”+\sin \left( y’ \right)+y=0\ \text{at}\ \left( 2,ND \right)\end{array} \)

Formation of Differential Equation

Let y and x be the dependent and independent variables, respectively, for the equation where k is an arbitrary constant. Then, here is the step to form a differential equation:

Count the number of independent arbitrary constants (let if ‘n’)

The number of arbitrary constants will be equal to the order of the differential equation. (n)

Differentiate the given equation ’n’ times to eliminate arbitrary constants.

The above elimination equation will be a required differential equation.

For family of curves: 

\(\begin{array}{l}f\left( x,y,\alpha ,{{\alpha }_{2}}….{{\alpha }_{n}} \right)=0….(1)\end{array} \)

Where, α1, α2, …., αn are n different parameters. If we differentiate n times, we get the differential equation of the given equation (1).

For example, form a differential equation of all lines passing through the origin.


\(\begin{array}{l}y=mx\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{dy}{dx}=m\end{array} \)

∴ xdy – ydx = 0 is the answer.

Solution of Differential Equations

The solution of the differential equation is the relation between the variables involved, which satisfies the differential equation.

Types of solutions

1. General solution:

It contains as many arbitrary constants as the order of the differential equation.

2. Particular solution

The solution is obtained by giving particular values to the arbitrary constants in the general solution of the differential equation.

5 Methods for Solving the Differential Equation

  1. Solution by inspection
  2. Variable separable
  3. Homogeneous
  4. Linear differential equation
  5. General

Solution by Inspection

If the differential equation is of the form

\(\begin{array}{l}f({{f}_{1}}(x,\,y))\ d({{f}_{1}}(x,\,y))+\varphi ({{f}_{2}}(x,\,y))\ d({{f}_{2}}(x,\,y))+……=0,\end{array} \)
, then each term can be separately integrated.
The solution to a differential equation can be found using the inspection method. It is accompanied by memorising the following results.

\(\begin{array}{l}(i)\ d(x+y)=dx+dy\\ (ii)\ d(xy)=xdy+ydx\\ (iii)\ d(\frac{x}{y})=\frac{ydx-xdy}{y^{2}}\\ (iv)\ d(\frac{y}{x})=\frac{xdy-ydx}{x^{2}}\\ (v)\ d(\frac{x^{2}}{y})=\frac{2xydx-x^{2}dy}{y^{2}}\\ (vi)\ d(\frac{y^{2}}{x})=\frac{2xydy-y^{2}dx}{x^{2}}\\ (vii)\ d(\frac{x^{2}}{y^{2}})=\frac{2xy^{2}dx-2x^{2}ydy}{y^{4}}\\ (viii)\ d(\frac{y^{2}}{x^{2}})=\frac{2x^{2}ydy-2xy^{2}dx}{x^{4}}\\ (ix)d(tan^{-1}\frac{x}{y})=\frac{ydx-xdy}{x^{2}+y^{2}} \\ (x)\ d(tan^{-1}\frac{y}{x})=\frac{xdy-ydx}{x^{2}+y^{2}}\\(xi)\ d(ln[xy])=\frac{xdx+ydx}{xy}\\ (xii)\ d(\frac{x}{y})=\frac{ydx-xdy}{xy}\\ (xiii)\ d(\frac{1}{2}ln(x^{2}+y^{2}))=\frac{xdx+ydy}{x^{2}+y^{2}}\\ (xiv)\ d[ln \frac{y}{x}]=\frac{xdy-ydx}{xy}\\ (xv)\ d(\frac{-1}{xy})=\frac{xdy+ydx}{x^{2}y^{2}}\\ (xvi)\ d(\frac{e^{x}}{y})=\frac{ye^{x}dx-e^{x}dy}{y^{2}}\\ (xvii)\ d(\frac{e^{y}}{x})=\frac{xe^{y}dy-e^{y}dx}{x^{2}}\\ (xviii)\ d(x^{m}y^{n})=x^{m-1}y^{n-1}(mydx+nxdy)\\ (xix)\ d(\sqrt{x^{2}+y^{2}})=\frac{xdy-ydx}{x^{2}-y^{2}}\\ (xx)\ d(\frac{1}{2}log\frac{x+y}{x-y})=\frac{xdy-ydx}{x^{2}-y^{2}}\\\end{array} \)

Variable Separable Method

If an equation can be written such that variables are separated for integration, then the equation can be solved.

We get

\(\begin{array}{l}\int{f\left( x \right)dx+\int{g\left( y \right)dy=c}}\end{array} \)

Homogeneous Differential Equation

A differential equation of form

\(\begin{array}{l}\frac{dy}{dx}=\frac{f\left( x,y \right)}{\phi \left( x,y \right)}\end{array} \)
, where f and Φ are homogeneous is called a homogeneous differential equation.

Homogeneous function: The function (x, y) is called a function. So, if

\(\begin{array}{l}f\left( \lambda x,\lambda y \right)={{\lambda }^{n}}f\left( x,y \right)\end{array} \)

Thus homogeneous function can be written as

\(\begin{array}{l}f\left( x,y \right)={{x}^{n}}f\left( \frac{y}{x} \right)\end{array} \)
\(\begin{array}{l}f\left( x,y \right)={{y}^{n}}f\left( \frac{y}{x} \right)\end{array} \)

Homogeneous differential equations

In the first order, the first degree differential equation is expressed in the form

\(\begin{array}{l}\frac{dy}{dx}=\frac{f\left( x,y \right)}{g\left( x,y \right)}\end{array} \)

Example: Solve differential equation x2 dy + y(x + y)dx = 0; y = 1 when x = 1.


\(\begin{array}{l}{{x}^{2}}dy+y\left( x+y \right)dx=0\end{array} \)
\(\begin{array}{l}{{x}^{2}}dy=-y\left( x+y \right)dx\end{array} \)
\(\begin{array}{l}\frac{dy}{dx}=\frac{-y\left( x+y \right)}{{{x}^{2}}}\end{array} \)

Since each xy + y2 and x2 are homogeneous,


\(\begin{array}{l}y=vx\end{array} \)
\(\begin{array}{l}\frac{dy}{dx}=v+x\frac{dv}{dx}\end{array} \)
\(\begin{array}{l}v+x\frac{dv}{dx}=-\left( \frac{v{{x}^{2}}+{{v}^{2}}{{x}^{2}}}{{{x}^{2}}} \right)\end{array} \)
\(\begin{array}{l}v+x\frac{dv}{dx}=-\left( v+{{v}^{2}} \right)\end{array} \)
\(\begin{array}{l}\int{\frac{1}{{{v}^{2}}+2v}dv=-\int{\frac{1}{x}}}\end{array} \)
\(\begin{array}{l}\frac{1}{2}\log \left| \frac{v+1-1}{v+1+1} \right|=-\log x+\log c\end{array} \)
\(\begin{array}{l}\log \left| \frac{v}{v+2} \right|+2\log x=2\log c\end{array} \)
\(\begin{array}{l}\log \left| \frac{v{{x}^{2}}}{v+2} \right|=2\log k\end{array} \)
\(\begin{array}{l}k=\left| \frac{v{{x}^{2}}}{v+2} \right|\end{array} \)
\(\begin{array}{l}k=\left| \frac{{{x}^{2}}y}{y+2x} \right|\end{array} \)

Put y = 1 & x = 1

k = 1/3

3x2 = y + 2x

\(\begin{array}{l}y=\frac{2x}{3{{x}^{2}}-1}\ \text{which is a required solution}.\end{array} \)

Example: Solve

\(\begin{array}{l}{{x}^{2}}ydx-\left( {{x}^{3}}+{{y}^{3}} \right)dy=0\end{array} \)


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

Putting y = vx and

\(\begin{array}{l}\frac{dy}{dx}=v+x\frac{dv}{dx},\end{array} \)
\(\begin{array}{l}v+x\frac{dv}{dx}=\frac{v{{x}^{3}}}{{{x}^{3}}+{{v}^{3}}{{x}^{3}}}\end{array} \)
\(\begin{array}{l}= \frac{v{x}^{3}}{{x}^{3}\left( 1+{v}^{3} \right)}\end{array} \)
\(\begin{array}{l}=\frac{v}{\left( 1+{{v}^{3}} \right)}\end{array} \)


\(\begin{array}{l}x\frac{dv}{dx}=\frac{v}{1+{{v}^{3}}}-v\end{array} \)
\(\begin{array}{l}x\left( 1+{{v}^{3}} \right)dv=-{{v}^{4}}dx\end{array} \)
\(\begin{array}{l}\left( \frac{1+{{v}^{3}}}{{{v}^{4}}} \right)dv=\frac{-dx}{x}\end{array} \)
\(\begin{array}{l}-\frac{1}{3{{v}^{3}}}+\log \left| v \right|+\log \left| x \right|=c\end{array} \)
\(\begin{array}{l}-\frac{1}{3}\frac{{{x}^{3}}}{{{y}^{3}}}+\log \left| \frac{y}{x}.\,x \right|=c\end{array} \)
\(\begin{array}{l}\frac{-{{x}^{3}}}{3{{y}^{3}}}+\log \left| y \right|=c\ \text{is the required solution.}\end{array} \)

We solve by y = tx, and proceed accordingly.

Linear Differential Equation

Equation of form (dy/dx) + py = θ.

Linear differential equation: A differential equation is linear if the dependent variable (y) and its derivative appear only in the first degree.

The general form of linear differential equation of first order is

\(\begin{array}{l}\frac{dy}{dx}+Py=Q\end{array} \)

Here, P and Q are constants.

\(\begin{array}{l}\text{We solve such type of equation by multiplying both sides}\ {{e}^{\int{Pdx}}},\end{array} \)


\(\begin{array}{l}{{e}^{\int{Pdx}}}\left( \frac{dy}{dx}+Py \right)=Q\,{{e}^{\int{Pdx}}}\end{array} \)
\(\begin{array}{l}\frac{d}{dx}\left\{ y\,{{e}^{\int{Pdx}}} \right\}=Q\,{{e}^{\int{Pdx}}}\end{array} \)

Integrating both sides,

\(\begin{array}{l}y\,{{e}^{\int{Pdx}}}=\int{\,Q\,\,{{e}^{\int{Pdx}}}dx+c}\end{array} \)


\(\begin{array}{l}{{e}^{\int{Pdx}}}\ \text{is called integrating factor.}\end{array} \)
\(\begin{array}{l}y[I.F]=\int{Q\,(I.F)dx+c}\end{array} \)

Applications of Differential Equations

  • For finding rate measurer
  • For finding the tangent to a curve


Illustration 1: If the population of a bacteria get doubles in 10 years, in how many years will it triple? Given that rate of growth is proportional to the number of bacteria.

Solution: Let the total number of bacteria be ‘x.’

\(\begin{array}{l}\therefore \frac{dx}{dt}\propto x\end{array} \)

\(\begin{array}{l}\therefore \frac{dx}{dt}=kx\end{array} \)

\(\begin{array}{l}\int{\frac{dx}{x}}=\int{k\,\,dt}\end{array} \)

\(\begin{array}{l}\log x=kt+C\end{array} \)

\(\begin{array}{l}x={{e}^{kt+C}}\Rightarrow \,\,x={{x}_{0}}{{e}^{kt}}\end{array} \)

Where (eC = x0) is population at t = 0 at t = 10 year, x = 2x0

\(\begin{array}{l}So,\,2{{x}_{0}}={{x}_{0}}\,\,{{e}^{k\times 10}}\end{array} \)

\(\begin{array}{l}k=\frac{\log 2}{10}\end{array} \)

Now, for x = 3x0

t = ?

\(\begin{array}{l}3{{x}_{0}}={{x}_{0}}.\,\,{{e}^{\left( \frac{\log 2}{10} \right)t}}\Rightarrow t=10\frac{\log 3}{\log 2}\end{array} \)

t = 15.9 years.

Illustration 2: Biologist, at t = 0, puts 100 bacteria in a favourable growth medium; 6 hours later, it became 450. Assume exponential growth find growth const k.


\(\begin{array}{l}A=P{{e}^{kt}}\end{array} \)


\(\begin{array}{l}450=100{{e}^{6k}}\end{array} \)

\(\begin{array}{l}k=\frac{\ell n\,4.5}{6}\end{array} \)

[1st order differential equations: dN/dt = kN]

Illustration 3:  5 mice in a stable population of 500 are infected with a disease to test a theory which says the rate of change of the infected population is α product of the diseased one and disease free.

How long it takes for half the population to contract the disease?


\(\begin{array}{l}\ell n\left( \frac{N}{500-N} \right)=500\left( kt+c \right)\end{array} \)


\(\begin{array}{l}\frac{N}{500-N}={{c}_{1}}\,\,{{e}^{500kt}}\,\,\, at\ t=0\,\,\,N=5\end{array} \)


\(\begin{array}{l}\frac{5}{495}={{c}_{1}}{{e}^{500k\left( 0 \right)}}\Rightarrow {{c}_{1}}=\frac{1}{99}\end{array} \)
\(\begin{array}{l}\therefore \frac{N}{500-N}=\frac{1}{99}\,{{e}^{500kt}}\end{array} \)


\(\begin{array}{l}N=250\Rightarrow 1=\frac{1}{99}{{e}^{500k{{T}_{1/2}}}}\end{array} \)


\(\begin{array}{l}{{T}_{1/2}}=\frac{\ell n\left( 99 \right)}{500k}=\frac{0.0091}{k}\end{array} \)
time units.

Newton’s cooling law

According to Newton’s cooling law, the rate of change in the temperature of a body is proportional to the difference in temperature of the body and its surroundings.

\(\begin{array}{l}\frac{dT}{dt}=-k\left( T-{{T}_{m}} \right)\end{array} \)

here Tm Surrounding temp

Cooling produces dT/dt → -ve,  T greater than Tm.

Illustration 5: A metal bar at 100°F at room temperature 0°F if, after 20 minutes, the temperature of the bar is 50°F.

(a) Time to reach 25°F

(b) Temp after 10 minutes


\(\begin{array}{l}\frac{dT}{dt}=-k\left( T-{{T}_{m}} \right)\Rightarrow \frac{dT}{dt}+kT=0\end{array} \)
\(\begin{array}{l}{{T}_{m}}=0\end{array} \)


\(\begin{array}{l}T=c{{e}^{-kt}}\end{array} \)

T = 100 at t = 0

\(\begin{array}{l}\therefore c=100\,{{e}^{k0}}\end{array} \)

\(\begin{array}{l}T=100\,{{e}^{-kt}}\end{array} \)

At t = 20 and T = 50

\(\begin{array}{l}\therefore \frac{1}{2}={{e}^{-k\,20}}\,\,\,\,\,\frac{\ell n2}{20}=k\end{array} \)



\(\begin{array}{l}25=100\,\,{{e}^{-\frac{\ell n2}{20}t}}\end{array} \)
\(\begin{array}{l}\ell n\,4=\frac{\ell {{n}^{2}}}{20}t\end{array} \)

or t = 40


\(\begin{array}{l}T=100{{e}^{-\frac{\ell n2}{20}\times 10}}=\frac{100}{\sqrt{2}}=70.71{}^\circ F\end{array} \)

First Order Differential Equations

Below are some of the most important and popular methods to find the solution to first-order and first-degree differential equations, along with examples.

Methods of solving

1. Differential equation is the form of dy/dx = f(x). In this, we integrate both sides to get general solutions.


\(\begin{array}{l}\frac{dy}{dx}=\frac{x}{{{x}^{2}}+1}\end{array} \)


\(\begin{array}{l}dy=\left( \frac{x}{{{x}^{2}}+1} \right)dx\end{array} \)


\(\begin{array}{l}\int{dy}=\int{\frac{xdx}{{{x}^{2}}+1}}\end{array} \)


\(\begin{array}{l}\int{dy}=\frac{1}{2}\int{\frac{2x\,\,dx}{{{x}^{2}}+1}}\end{array} \)


\(\begin{array}{l}y=\frac{1}{2}\log \left| {{x}^{2}}+1 \right|+c;\,\,\,\,\,\,\,\,x\in R\end{array} \)
is the solution of given differential equation.


\(\begin{array}{l}\frac{dy}{dx}=\frac{3{{e}^{2x}}+3{{e}^{4x}}}{{{e}^{x}}+{{e}^{-x}}}\end{array} \)


\(\begin{array}{l}\frac{dy}{dx}=\frac{3{{e}^{2x}}+3{{r}^{4x}}}{{{e}^{x}}+\frac{1}{{{e}^{x}}}}=\frac{3{{e}^{2x}}\left( 1+{{e}^{2x}} \right)}{{{e}^{x}}+\frac{1}{{{e}^{x}}}}\end{array} \)


\(\begin{array}{l}\frac{dy}{dx}=3{{e}^{3x}}\end{array} \)


\(\begin{array}{l}\int{dy}=\int{3{{e}^{3x}}dx}\end{array} \)


\(\begin{array}{l}y=3\left( \frac{{{e}^{3x}}}{3} \right)+c\end{array} \)
is required solutions.

2. Differential equations reducible to variable separable type.

When the first-order differential equation cannot be solved by integration, then we apply the substitution method in order to reduce the variable.

A differential equation of the form

\(\begin{array}{l}\frac{dy}{dx}=f\left( px+qy \right)\end{array} \)

Substitute px + qy = t

For example, solve

\(\begin{array}{l}\frac{dy}{dx}=\left( ax+by \right)\end{array} \)



\(\begin{array}{l}ax+by=t\end{array} \)

\(\begin{array}{l}a+b\frac{dy}{dx}=\frac{dt}{dx}\end{array} \)

\(\begin{array}{l}\frac{dy}{dx}=\frac{1}{b}\left( \frac{dt}{dx}-a \right)\end{array} \)

\(\begin{array}{l}\frac{1}{b}\left( \frac{dt}{dx}-a \right)=t\end{array} \)

\(\begin{array}{l}\frac{dt}{dx}=bt+a\end{array} \)

\(\begin{array}{l}\int{dx}=\frac{1}{b}\int{\frac{bdt}{bt+a}}\end{array} \)

\(\begin{array}{l}x=\frac{1}{b}\ln \left( bt+a \right)\end{array} \)

\(\begin{array}{l}x=\frac{1}{b}\ln \left( b\left( ax+by \right)+a \right)\end{array} \)

3. First-order differential equation of homogeneous form:

Differential equation of the form:

\(\begin{array}{l}\frac{dy}{dx}=g\left( x,y \right)\end{array} \)

\(\begin{array}{l}\frac{dy}{dx}=f\left( y/x \right)\end{array} \)
are called homogeneous form.

For such equations, substitute y/x = t

y = xt

\(\begin{array}{l}\Rightarrow \frac{dy}{dx}=t+x\frac{dt}{dx}\end{array} \)

\(\begin{array}{l}\Rightarrow t+x\frac{dt}{dx}=f\left( t \right)\end{array} \)

\(\begin{array}{l}\Rightarrow x\frac{dt}{dx}=f\left( t \right)-t\end{array} \)

\(\begin{array}{l}\Rightarrow \int{\frac{dt}{f\left( t \right)-t}}=\int{\frac{dx}{x}}\end{array} \)

And then, ‘t’ can be replaced with (y/x) to get a solution of the differential equation.

For example, solve the differential equation

\(\begin{array}{l}\frac{dy}{dx}=\frac{3x+y}{x-y}\end{array} \)


\(\begin{array}{l}\frac{dy}{dx}=g\left( x,y \right)=\frac{3x+y}{x-y}\end{array} \)

\(\begin{array}{l}=f\left( \frac{y}{x} \right)=\frac{3+\frac{y}{x}}{1-\frac{y}{x}}\end{array} \)

\(\begin{array}{l}\frac{y}{x}=t\end{array} \)

\(\begin{array}{l}y=xt\,\,\,\,\,\Rightarrow f\left( t \right)=\frac{3+t}{1-t}\end{array} \)

\(\begin{array}{l}\frac{dy}{dx}=t+x\frac{dt}{dx}=f\left( t \right)\end{array} \)

\(\begin{array}{l}\Rightarrow \,\,\,x\frac{dt}{dx}=f\left( t \right)-t\end{array} \)

\(\begin{array}{l}=\frac{3+3t}{1-t}-t\end{array} \)

\(\begin{array}{l}=\frac{3+t-t+{{t}^{2}}}{1-t}\end{array} \)

\(\begin{array}{l}x\frac{dt}{dx}=\frac{3+{{t}^{2}}}{1-t}\end{array} \)

\(\begin{array}{l}\int{\frac{dx}{x}}=\int{\frac{\left( 1-t \right)dt}{3+{{t}^{2}}}}\end{array} \)

\(\begin{array}{l}\ln x+{{C}_{1}}=\int{\frac{dt}{3+{{t}^{2}}}}-\frac{1}{2}\int{\frac{2t}{3+{{t}^{2}}}}\,dt\end{array} \)

\(\begin{array}{l}=\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{t}{\sqrt{3}} \right)-\frac{1}{2}\ln \left( 3+{{t}^{2}} \right)+{{C}_{2}}\end{array} \)

\(\begin{array}{l}\ln x+{{C}_{1}}=\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{y}{\sqrt{3}x} \right)-\frac{1}{2}\ln \left( 3+\frac{{{y}^{2}}}{{{x}^{2}}} \right)+{{C}_{2}}\end{array} \)

Practice Problems on Differential Equations

Example 1: Solve

\(\begin{array}{l}\frac{dy}{dx}+y=1\end{array} \)


\(\begin{array}{l}\frac{dy}{dx}=1-y\end{array} \)


\(\begin{array}{l}\frac{dx}{dy}=\frac{1}{1-y}\end{array} \)


\(\begin{array}{l}\int{dx}=\int{\frac{dy}{1-y}}\end{array} \)

x = – log |1 – y| + c is the required solution.

Example 2: Solve

\(\begin{array}{l}\frac{dy}{dx}=\sec y\end{array} \)


\(\begin{array}{l}\frac{dx}{dy}=\frac{1}{\sec y}=\cos y\end{array} \)


\(\begin{array}{l}\int{dx}=\int{\cos \,y\,dy}\end{array} \)

x = sin y + c is the required solution.

Example 3: Solve

\(\begin{array}{l}\left( x+1 \right)\frac{dy}{dx}=2xy\end{array} \)


\(\begin{array}{l}\left( x+1 \right)dy=2xy\,\,dx\end{array} \)


\(\begin{array}{l}\frac{dy}{y}=\frac{2x\,dx}{x+1}\end{array} \)


\(\begin{array}{l}\frac{dy}{y}=\frac{2x\,\,dx}{x+1}\end{array} \)


\(\begin{array}{l}\int{\frac{dy}{y}}=\int{\frac{2x\,\,dx}{\left( x+1 \right)}}\end{array} \)
\(\begin{array}{l}\log \,y=2\left\{ x-\log \left| x+1 \right| \right\}+c\ \text{is the solutions.}\end{array} \)

Example 4: Solve

\(\begin{array}{l}{{e}^{x}}\sqrt{1-{{y}^{2}}\,}dx+\frac{y}{x}dy=0\end{array} \)


\(\begin{array}{l}x{{e}^{x}}dx=\frac{-y}{\sqrt{1-{{y}^{2}}}}dy\end{array} \)

Apply product rule of integration

\(\begin{array}{l}x{{e}^{x}}-\int{{{e}^{x}}}dx=\frac{1}{2}\int{\frac{dt}{\sqrt{t}}}t=1-{{y}^{2}}\end{array} \)
\(\begin{array}{l}x{{e}^{x}}-{{e}^{x}}=\sqrt{t}+c\end{array} \)
\(\begin{array}{l}x{{e}^{x}}-{{e}^{x}}=\sqrt{1+{{y}^{2}}}+c\ \text{is the required solutions.}\end{array} \)

Example 5: Solve

\(\begin{array}{l}\frac{dy}{dx}={{e}^{x+y}}\end{array} \)


\(\begin{array}{l}\frac{dy}{dx}={{e}^{x}}{{e}^{y}}\end{array} \)
\(\begin{array}{l}dy={{e}^{x}}.{{e}^{y}}.dx\end{array} \)
\(\begin{array}{l}\frac{dy}{{{e}^{y}}}={{e}^{x}}\,dx\end{array} \)
\(\begin{array}{l}\int{{{e}^{-y}}}\,dy=\int{{{e}^{x}}}dx\end{array} \)
\(\begin{array}{l}\frac{{{e}^{-y}}}{-1}={{e}^{x}}+C\end{array} \)
\(\begin{array}{l}-{{e}^{-y}}={{e}^{x}}+C\end{array} \)

Example 6: Solve

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


x + y = v


\(\begin{array}{l}\frac{dy}{dx}+1=\frac{dv}{dx}\end{array} \)
\(\begin{array}{l}\frac{dv}{dx}=\frac{dy}{dx}+1\Rightarrow \frac{dv}{dx}-1=\frac{dy}{dx}\end{array} \)

Put these in D.E,

\(\begin{array}{l}{{V}^{2}}\left( \frac{dv}{dx}-1 \right)={{a}^{2}}\end{array} \)
\(\begin{array}{l}{{v}^{2}}\frac{dv}{dx}-{{v}^{2}}={{a}^{2}}\end{array} \)
\(\begin{array}{l}{{v}^{2}}\frac{dv}{dx}={{a}^{2}}+{{v}^{2}}\end{array} \)
\(\begin{array}{l}\int{\left( \frac{{{V}^{2}}}{{{a}^{2}}+{{v}^{2}}} \right)}dv=\int{dx}\end{array} \)
\(\begin{array}{l}v-a\,{{\tan }^{-1}}\left( \frac{v}{a} \right)=x+c\end{array} \)
\(\begin{array}{l}\left( x+y \right)-a{{\tan }^{-1}}\left( \frac{x+y}{a} \right)=x+C\ \text{is the required solution.}\end{array} \)

Example 7: Solve

\(\begin{array}{l}\frac{dy}{dx}-\frac{y}{x}=2{{x}^{2}}\end{array} \)


\(\begin{array}{l}\frac{dy}{dx}+\left( -\frac{1}{x} \right)y=2{{x}^{2}}\end{array} \)
\(\begin{array}{l}P=\frac{-1}{x}\,\,and\,\,Q=2{{x}^{2}}\end{array} \)

Multiply both sides by I.F

\(\begin{array}{l}I.F.={{e}^{\int{Pdx}}}={{e}^{\int{\frac{-1}{x}dx}}}={{e}^{-\log x}}\end{array} \)
\(\begin{array}{l}={{e}^{\log {{x}^{-1}}}}\end{array} \)
\(\begin{array}{l}={{x}^{-1}}=\frac{1}{x}\end{array} \)
\(\begin{array}{l}\frac{1}{x}\frac{dy}{dx}-\frac{1}{{{x}^{2}}}y=2x\end{array} \)

Integrating both sides w.r.t. x, we get;

\(\begin{array}{l}y\left( \frac{1}{x} \right)=\int{2x\,dx+c}\end{array} \)
\(\begin{array}{l}\frac{y}{x}={{x}^{2}}+c\end{array} \)
\(\begin{array}{l}y={{x}^{3}}+cx\ \text{is required solution.}\end{array} \)

Differential Equations – JEE Advanced PY Questions

Differential Equations – Important Topics

Differential-Equations - Important Topics

Differential Equations – Important Questions

Differential-Equations - Important Questions

Differential Calculus

Differential Calculus

Top JEE Advanced Questions from Differential Equations

Frequently Asked Questions


Give an example of a differential equation.

An example of a differential equation is (d2y/dx2) – (dy/dx) – 3 = 0.


Define the order of a differential equation.

The order of a differential equation is the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation.


What do you mean by the degree of a differential equation?

The power of the highest order derivative present in the differential equation is the degree of the differential equation.


Give two applications of differential equations.

Newton’s law of fall of an object and Newton’s law of cooling are two applications of differential equations.


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