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Introduction to Differential Equations

An equation which involves derivatives of a dependent variable with respect to other 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 medical, chemistry, physics and engineering. It is a tool which helps in building 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. Differential equations is 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 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 highest order derivative
\(\begin{array}{l}f\left( xy \right)\frac{{{d}^{m}}y}{d{{x}^{m}}}+g\left( xm \right)\frac{{{d}^{m-1}}y}{d{{x}^{m-1}}}+…=0\end{array} \)

Note:

Degree of differential equation is only define when it is in polynomial form.

Degree may not be defined at ways:

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

Formation of Differential Equation

Let y and x be the dependent and independent variable respectively for equation  where k is arbitrary constant. Now, step to form differential equation:

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

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

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

Above elimination equation will be required differential equation.

For family of curves: 

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

Where

\(\begin{array}{l}\alpha ,{{\alpha }_{2}}…{{\alpha }_{n}}\end{array} \)
are n different parameters. If we differentiate n times, we get differential equation of the given equation (1).

For example, Form differential equation of all lines passing through origin.

Answer:

\(\begin{array}{l}y=mx\end{array} \)

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

∴ xdy – ydx = 0 is answer.

Solution of Differential Equations

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

Types of solutions:

1. General solution:

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

2. Particular solution

The solution obtained by giving particular values to the arbitrary constants in the general solution of 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

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

Homogenous 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
\(\begin{array}{l}f\And \phi\end{array} \)
are homogeneous is called as homogenous.

Homogenous function: The function (x, y) is called functions. So, if

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

Thus homogenous function can be written as

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

Homogeneous differential equations

In first order first degree differential equation is expressed in form.

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

Example: Solve differential equation

\(\begin{array}{l}{{x}^{2}}dy+y\left( x+y \right)dx=0\end{array} \)
; y = 1 when x = 1.

Solution:

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

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

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

Since each xy + y2 and x2 are homogeneous.

Putting,

\(\begin{array}{l}y=vx\end{array} \)
,

\(\begin{array}{l}\frac{dy}{dx}=v+x\frac{dv}{dx}\end{array} \)
,

\(\begin{array}{l}v+x\frac{dv}{dx}=-\left( \frac{v{{x}^{2}}+{{v}^{2}}{{x}^{2}}}{{{x}^{2}}} \right)\end{array} \)
,

\(\begin{array}{l}v+x\frac{dv}{dx}=-\left( v+{{v}^{2}} \right)\end{array} \)
,

\(\begin{array}{l}\int{\frac{1}{{{v}^{2}}+2v}dv=-\int{\frac{1}{x}}}\end{array} \)
,

\(\begin{array}{l}\frac{1}{2}\log \left| \frac{v+1-1}{v+1+1} \right|=-\log x+\log c\end{array} \)
,

\(\begin{array}{l}\log \left| \frac{v}{v+2} \right|+2\log x=2\log c\end{array} \)
,

\(\begin{array}{l}\log \left| \frac{v{{x}^{2}}}{v+2} \right|=2\log k\end{array} \)
,

\(\begin{array}{l}k=\left| \frac{v{{x}^{2}}}{v+2} \right|\end{array} \)
,

\(\begin{array}{l}k=\left| \frac{{{x}^{2}}y}{y+2x} \right|\end{array} \)

Put y = 1 & x = 1

\(\begin{array}{l}k=\frac{1}{3}\end{array} \)
,

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

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

Example: Solve

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

Solution:

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

Putting y = vx and

\(\begin{array}{l}\frac{dy}{dx}=v+x\frac{dv}{dx}\end{array} \)
,

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

=

\(\begin{array}{l}\frac{v{x}^{3}}{{x}^{3}\left( 1+{v}^{3} \right)}\end{array} \)

=

\(\begin{array}{l}\frac{v}{\left( 1+{{v}^{3}} \right)}\end{array} \)

or

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

\(\begin{array}{l}x\left( 1+{{v}^{3}} \right)dv=-{{v}^{4}}dx\end{array} \)
,

\(\begin{array}{l}\left( \frac{1+{{v}^{3}}}{{{v}^{4}}} \right)dv=\frac{-dx}{x}\end{array} \)
,

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

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

\(\begin{array}{l}\frac{-{{x}^{3}}}{3{{y}^{3}}}+\log \left| y \right|=c\end{array} \)
is the required solution.

We solve by

\(\begin{array}{l}y=tx\end{array} \)
& proceeding accordingly.

Linear Differential Equation

Equation of form

\(\begin{array}{l}\frac{dy}{dx}+py=\theta\end{array} \)
.

Linear differential equation: A differential equation is linear if the dependent variable (y) and its derivative appear only in 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, Q are constants.

We solve such type of equation by multiplying both sides

\(\begin{array}{l}{{e}^{\int{Pdx}}}\end{array} \)
, so

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

Integrating both sides

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

Here,

\(\begin{array}{l}{{e}^{\int{Pdx}}}\end{array} \)
is called integrating factor.

\(\begin{array}{l}y[I.F]=\int{Q\,(I.F)dx+c}\end{array} \)

Applications of Differential Equations

Applications of differential equation

Illustrations:

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

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

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

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

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

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

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

Where

\(\begin{array}{l}\left( {{e}^{C}}={{x}_{0}} \right)\end{array} \)
is population at t = 0 at t = 10 year,
\(\begin{array}{l}x=2{{x}_{0}}\end{array} \)

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

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

Now, for

\(\begin{array}{l}x=3{{x}_{0}}\end{array} \)

\(\begin{array}{l}t=?\end{array} \)

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

\(\begin{array}{l}t=15.9\end{array} \)
years

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

Solution:

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

or

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

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

[1st order differential equations: 

\(\begin{array}{l}\frac{dN}{dt}=kN\end{array} \)
]

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

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

Solution:

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

or

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

or

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

\(\begin{array}{l}\frac{N}{500-N}=\frac{1}{99}\,{{e}^{500kt}}\end{array} \)

or

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

or

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

Newton’s cooling law

Time rate of change of temp of a body is α to temp difference between body & its surrounding medium

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

here

\(\begin{array}{l}{{T}_{m}}\to\end{array} \)
Surrounding temp

Cooling produces

\(\begin{array}{l}\frac{dT}{dt}\to -\end{array} \)
ve,  T greater than Tm.

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

(a) Time to reach 25°F

(b) Temp after 10 min.

Solution:

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

or

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

T = 100 at t = 0

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

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

At t = 20 and T = 50

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

Now,

(a)

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

or t = 40

(b)

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

First Order Differential Equations

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

Methods of solving

1. Differential equation is form of

\(\begin{array}{l}\frac{dy}{dx}=f\left( x \right)\end{array} \)
. In this we integrate both sides to get general solutions.

Example:

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

Solution:

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

or

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

or

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

or

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

Example:

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

Solution:

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

or

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

or

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

or

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

2. Differential equations reducible to variable separable type.

When first order differential equation could not solved by integration, then we apply 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

\(\begin{array}{l}px+qy=t\end{array} \)

For example: Solve

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

Solution:

Assume

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

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

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

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

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

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

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

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

3. First order differential equation of Homogeneous form:

Differential equation of the form:

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

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

For such equations substitutes

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

\(\begin{array}{l}y=xt\end{array} \)

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

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

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

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

And then ‘t’ can be replaced with

\(\begin{array}{l}\left( \frac{y}{x} \right)\end{array} \)
to get a solution of differential equation.

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

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

Solution:

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

or

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

or

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

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

 

Example 2: Solve

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

Solutions:

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

or

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

x = sin y + c is required solutions.

Example 3: Solve

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

Solutions:

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

or

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

or

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

or

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

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

Example 4: Solve

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

Solutions:

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

Apply product rule of integration

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

\(\begin{array}{l}x{{e}^{x}}-{{e}^{x}}=\sqrt{t}+c\end{array} \)
,

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

Example 5: Solve

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

Solution:

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

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

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

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

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

\(\begin{array}{l}-{{e}^{-y}}={{e}^{x}}+C\end{array} \)

Example 6: Solve

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

Solution:

x + y = v

Then

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

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

Put these in D.E,

\(\begin{array}{l}{{V}^{2}}\left( \frac{dv}{dx}-1 \right)={{a}^{2}}\end{array} \)
,

\(\begin{array}{l}{{v}^{2}}\frac{dv}{dx}-{{v}^{2}}={{a}^{2}}\end{array} \)
,

\(\begin{array}{l}{{v}^{2}}\frac{dv}{dx}={{a}^{2}}+{{v}^{2}}\end{array} \)
,

\(\begin{array}{l}\int{\left( \frac{{{V}^{2}}}{{{a}^{2}}+{{v}^{2}}} \right)}dv=\int{dx}\end{array} \)
,

\(\begin{array}{l}v-a\,{{\tan }^{-1}}\left( \frac{v}{a} \right)=x+c\end{array} \)
,

\(\begin{array}{l}\left( x+y \right)-a{{\tan }^{-1}}\left( \frac{x+y}{a} \right)=x+C\end{array} \)
is a required solutions.

Example 7: Solve

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

Solution:

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

\(\begin{array}{l}P=\frac{-1}{x}\,\,and\,\,Q=2{{x}^{2}}\end{array} \)

Multiply both sides by I.F

I.F

\(\begin{array}{l}={{e}^{\int{Pdx}}}={{e}^{\int{\frac{-1}{x}dx}}}={{e}^{-\log x}}\end{array} \)
,

\(\begin{array}{l}={{e}^{\log {{x}^{-1}}}}\end{array} \)
,

\(\begin{array}{l}={{x}^{-1}}=\frac{1}{x}\end{array} \)
,

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

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

\(\begin{array}{l}y\left( \frac{1}{x} \right)=\int{2x\,dx+c}\end{array} \)
,

\(\begin{array}{l}\frac{y}{x}={{x}^{2}}+c\end{array} \)
,

\(\begin{array}{l}y={{x}^{3}}+cx\end{array} \)
is required solution.

Differential Equations – Important Topics

Differential-Equations - Important Topics

Differential Equations – Important Questions

Differential-Equations - Important Questions

Differential Calculus

Differential Calculus

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 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.