Informative line

### Different Types Of Differential Equations And Their Solutions

Learn solution differential equation to Integrating Factor and practice Integral, Homogeneous, Linear, and Bernoulli differential equations.

# Solving an Integral Equation

## Integral equation

• An integral equation is an equation having a relation between an unknown function $$y(x)$$ and an integral containing $$y(x)$$.

For example: $$\displaystyle y(x)=\int\limits^x_3(y(t)+t)\,dt$$

### Solution of integral equation

• To solve these equations, we use Leibnitz's rule.
• The Leibnetz's rule is $$\displaystyle \dfrac{d}{dx}\int\limits^x_a(y(t)+t)=f(x)$$

i.e., if we integrate a function '$$f$$' and then differentiate it, we get the same function  back.

• As a result, we will obtain a differential equation which can be easily solved by variable separate method.

Example:

$$\displaystyle y(x)=\int\limits^x_3[y(t)+t]\,dt$$

Differentiating both sides with respect to $$x$$

$$\displaystyle \dfrac{dy(x)}{dx}=\dfrac{d}{dx}\int\limits^x_3[y(t)+t]\,dt$$

use Leibnitz's rule,

$$\dfrac{dy(x)}{dx}=y(x)+x$$

$$dy(x)=(y(x)+x)\,dx$$

Integrating on both sides

$$\int dy(x)=\int[y(x)+x]\,dx$$

We will get the required solution.

### particular solution of integral equation

• To get the particular solution, put the value of $$x$$ in the solution obtained (by Leibnitz's rule).
• Then calculate the value of constant.
• Put the value of constant in general solution and obtain the particular solution.

#### Solve the following integral equation $$\displaystyle y(x)=5+\int\limits^x_2(t^2-t^2y(t))dt$$

A $$y=7x^2+x$$

B $$y=2sinx+lnx$$

C $$y=8e^{-x/2}$$

D $$|1-y|=4e^\left(\dfrac{8-x^3}{3}\right)$$

×

$$\displaystyle y(x)=5+\int\limits^x_2(t^2-t^2y(t))dt$$ is the given integral equation.

Differentiate both sides with respect to  $$x$$

$$\Rightarrow\,y'(x)=0+(x^2-x^2y(x))\to$$Use Leibniz's rule

$$\Rightarrow\,y'(x)=x^2(1-y(x))\Rightarrow\,\dfrac{dy}{dx}=x^2(1-y)$$

The equation is of variable separable form

$$\therefore\,\displaystyle \int\dfrac{1}{1-y}dy=\int x^2dx\Rightarrow-ln|1-y|=\dfrac{x^3}{3}+C$$

Form the integral equation

$$\displaystyle y(2)=5+\int\limits^2_2(t^2-t^2y(t))dt$$

$$\Rightarrow\,y(2)=5+0=5\Rightarrow y(2)=5$$

Use this in general solution, when $$x=2,\,y=5$$

$$-ln|1-5|=\dfrac{8}{3}+C\Rightarrow\,-ln4=\dfrac{8}{3}+C$$

$$\Rightarrow\,C=\dfrac{-8}{3}-ln4.$$

$$\therefore$$ Particular solution is

$$-ln|1-y|=\dfrac{x^3}{3}-\dfrac{8}{3}-ln4$$

$$\Rightarrow\,ln\dfrac{|1-y|}{4}=\dfrac{8-x^3}{3}$$

$$\Rightarrow|1-y|=4e^\left(\dfrac{8-x^3}{3}\right)$$

### Solve the following integral equation $$\displaystyle y(x)=5+\int\limits^x_2(t^2-t^2y(t))dt$$

A

$$y=7x^2+x$$

.

B

$$y=2sinx+lnx$$

C

$$y=8e^{-x/2}$$

D

$$|1-y|=4e^\left(\dfrac{8-x^3}{3}\right)$$

Option D is Correct

# Homogeneous Differential Equation

A differential equation of the form

$$\dfrac{dy}{dx}=f\left(\dfrac{y}{x}\right){\to{(1)}}$$ is called a homogeneous differential equation

• The R.H.S function depends entirely on  $$\dfrac{y}{x}$$.
• First, arrange the differential equation in the form,

$$\dfrac{dy}{dx}=f\left(\dfrac{y}{x}\right)\;\text{or}\;\dfrac{dx}{dy}=f\left(\dfrac{x}{y}\right)$$

• If an equation gets arranged in the above form and each term of RHS of the equation has the same degree, the differential equation is a homogeneous differential equation.

Example : $$\dfrac{dy}{dx}=\dfrac{x^2+y^2}{xy}\dfrac{\longrightarrow\;deg=2}{\longrightarrow\;deg=2}$$

• In these differential equation we make the substitution.

$$y=vx\Rightarrow\,\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$$

$$\therefore$$ Equation (1) becomes

$$v+x\dfrac{dv}{dx}=f(v)$$

$$\Rightarrow\,x\dfrac{dv}{dx}=f(v)-v^{\nearrow^{\text{which is variable separable}}}$$

$$\Rightarrow\,\dfrac{dv}{f(v)-v}=\dfrac{dx}{x}$$

Now integrate both sides

$$\displaystyle\int\dfrac{dv}{f(v)-v}=ln\,cx$$

• So putting $$y=vx$$ , changes the equation to variable separable function.

Note: If the homogeneous differential equation is of the form,

$$\dfrac{dx}{dy}=f\left(\dfrac{x}{y}\right)$$

then make the substitution, $$\dfrac{x}{y}=v$$

$$\Rightarrow\;x=vy$$

and solve accordingly.

#### Solve $$\dfrac{dy}{dx}=\dfrac{y+x}{x}$$

A $$y=ce^x+1$$

B $$y=xln|cx|$$

C $$y=csin^2x+x$$

D $$y=x^2ln|cx|$$

×

Homogeneous differential equation one of the form

$$\dfrac{dy}{dx}=f\left(\dfrac{y}{x}\right)$$

Here  $$\dfrac{dy}{dx}=\dfrac{y}{x}+1\to$$ (1)

put $$y=vx\Rightarrow\,\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$$

$$\therefore$$ equation (1) becomes

$$v+x\dfrac{dv}{dx}=v+1\Rightarrow\,x\dfrac{dv}{dx}=1$$

$$\Rightarrow\,dv=\dfrac{dx}{x}\to \text{variable separable}$$

Integrate both sides

$$\displaystyle\int dv=\int\dfrac{dx}{x}\Rightarrow v=ln|cx|$$.

put $$v=\dfrac{y}{x}\Rightarrow\,\dfrac{y}{x}=ln|cx|$$

$$\Rightarrow\,y=xln|cx|\to \text{general solution}$$

### Solve $$\dfrac{dy}{dx}=\dfrac{y+x}{x}$$

A

$$y=ce^x+1$$

.

B

$$y=xln|cx|$$

C

$$y=csin^2x+x$$

D

$$y=x^2ln|cx|$$

Option B is Correct

#### Solve  $$(x\,ln\,x)\dfrac{dy}{dx}+y=\dfrac{2}{x}ln\,x$$

A $$y=\dfrac{-2}{x}-\dfrac{2}{x\,ln\,x}+\dfrac{c}{ln\,x}$$

B $$y=c\,sin^2x-cos\,x$$

C $$y=\dfrac{-2}{ln\,x}+\dfrac{3}{x}+c$$

D $$y=ce^{2x}+sin\,x$$

×

The first order linear differential equation is of the form

$$\dfrac{dy}{dx}+y\,P(x)=Q(x)$$

the integrating factor is $$I.F.=e^{\int P(x)dx}$$ and general solution is

$$y×I.F.=\int Q(x)×I.F.\,dx$$

In this case the differential equation is

$$x\,ln\,x\dfrac{dy}{dx}+y=\dfrac{2}{x}ln\,x$$

$$\dfrac{dy}{dx}+\dfrac{y}{x\,ln\,x}=\dfrac{2}{x^2}$$ which is linear

with $$P(x)=\dfrac{1}{x\,ln\,x},\,Q(x)=\dfrac{2}{x^2}$$.

$$I.F.=e^{\int P(x)dx}=e^{\int \dfrac{1}{x\,ln\,x}dx}\,$$

$$e^{ln(ln\,x)}=ln\,x$$

To integrated $$\displaystyle I={\int\dfrac{1}{x\,ln\,x}}dx$$

put  $$ln\,x=t$$

$$\Rightarrow\dfrac{1}{x}dx=dt$$

$$\displaystyle\Rightarrow\,I=\int \dfrac{1}{t}dt=ln\,t=ln(ln\,x)$$

$$\therefore\,I.F.=ln\,x$$

$$\therefore\,$$ General solution is

$$\displaystyle y×ln\,x=\int\dfrac{2}{x^2}ln\,x\,dx$$

Now  $$\displaystyle\int \dfrac{2}{x^2}ln\,x\,dx=2ln\,x×\dfrac{-1}{x}-\int\dfrac{2}{x}×\dfrac{-1}{x}dx$$

$$=\dfrac{-2ln\,x}{x}-\dfrac{2}{x}+c\,\,$$ (integration by parts)

$$\Rightarrow\,y\,ln\,x=\dfrac{-2ln\,x}{x}-\dfrac{2}{x}+c$$

$$\Rightarrow\,y=\dfrac{-2}{x}-\dfrac{2}{x\,ln\,x}+\dfrac{c}{ln\,x}$$

### Solve  $$(x\,ln\,x)\dfrac{dy}{dx}+y=\dfrac{2}{x}ln\,x$$

A

$$y=\dfrac{-2}{x}-\dfrac{2}{x\,ln\,x}+\dfrac{c}{ln\,x}$$

.

B

$$y=c\,sin^2x-cos\,x$$

C

$$y=\dfrac{-2}{ln\,x}+\dfrac{3}{x}+c$$

D

$$y=ce^{2x}+sin\,x$$

Option A is Correct

#### Solve the differential equation $$\dfrac{xdy}{dx}-y=lnx\,,\,y(1)=0$$

A $$y=xln\,x-2$$

B $$y=x-1-lnx$$

C $$y=2sinx-e^x$$

D $$y=x+x^2$$

×

The first order linear differential equation is of the form

$$\dfrac{dy}{dx}+y\,P(x)=Q(x)$$

the integrating factor is $$I.F.=e^{\int P(x)dx}$$ and general solution is

$$y×I.F.=\int Q(x)×I.F.\,dx$$

In this case the differential equation is

$$x\dfrac{dy}{dx}-y=ln\,x$$

$$\Rightarrow\,\dfrac{dy}{dx}-\dfrac{y}{x}=\dfrac{lnx}{x}$$ which is linear differential with

$$P(x)=\dfrac{-1}{x},\,Q(x)=\dfrac{lnx}{x}$$.

$$I.F.=e^{\int P(x)dx}$$

consider $$P(x)=\dfrac{-1}{x}$$

$$\displaystyle\Rightarrow\,{\int P(x)dx=\int\dfrac{-1}{x}dx}\,=-lnx$$

$$\therefore\,I.F.=e^{-lnx}$$

$$=e^{ln(x)^{-1}}=\dfrac{1}{x}$$  $$(\because\,e^{ln\,b}=b)$$

$$\therefore\,$$ General solution is

$$\displaystyle y×\dfrac{1}{x}=\int\dfrac{lnx}{x}×\dfrac{1}{x}dx$$

$$\Rightarrow \,\displaystyle \dfrac{y}{x}=\int\dfrac{lnx}{x^2}\,dx$$

$$\Rightarrow \,\displaystyle \dfrac{y}{x}=lnx×\dfrac{-1}{x}+\int\dfrac{1}{x}×\dfrac{1}{x}\,dx$$ (Use I.B.P.)

$$\Rightarrow\,\dfrac{y}{x}=\dfrac{-ln\,x}{x}-\dfrac{1}{x}+c$$

$$\Rightarrow\,y=-lnx-1+cx\to{\text{general solution}}$$

Now use initial condition give $$\to y(1)=0$$

i.e. when  $$x=1,\,y=0$$

$$\Rightarrow\,0=-ln1-1+c$$

$$\Rightarrow\,c=+1$$

$$\therefore$$ particular solution is

$$y=-lnx-1+x$$

$$\Rightarrow\,y=x-1-lnx$$

### Solve the differential equation $$\dfrac{xdy}{dx}-y=lnx\,,\,y(1)=0$$

A

$$y=xln\,x-2$$

.

B

$$y=x-1-lnx$$

C

$$y=2sinx-e^x$$

D

$$y=x+x^2$$

Option B is Correct

# Linear Differential Equation

• A first order linear differential equation is of the form

$$\dfrac{dy}{dx}+y\,P(x)=Q(x)$$

Where $$P(x)$$ and $$Q(x)$$ are functions of $$x$$ and are continuous.

• Linear equations occur frequently in various sciences and therefore the solution to these equation is very important.

## Identification of linear differential equation

• The first order linear differential equation can be identified easily as this equation does not contain the product of function with its derivative.

$$let\;y=f(x)$$

For example: $$\dfrac{dy}{dx}+xy=x^2\;(x\to\text{independent variable})$$

• Also, the power of function and its derivative is 1, separately.
• Note that these are not variable separable and will be solved by multiplying both sides by a factor called Integrating factor.

#### Which of the following differential equations is linear ?

A $$\dfrac{dy}{dx}+y\,lnx=x^2y$$

B $$\dfrac{dy}{dx}+y\,sinx=cosx$$

C $$x\dfrac{dy}{dx}+y^2\,cosx=5x$$

D $$\left(\dfrac{dy}{dx}\right)^2=5x$$

×

A linear differential equation is of the form

$$\dfrac{dy}{dx}+yP(x)=Q(x)$$

When $$P(x)$$ and $$Q(x)$$ are continuous function of $$x$$.

Only option 'b' is in this form with $$P(x)=sin\,x$$ and $$Q(x)=cos\,x$$

$$\therefore\,$$ option 'B' is correct.

### Which of the following differential equations is linear ?

A

$$\dfrac{dy}{dx}+y\,lnx=x^2y$$

.

B

$$\dfrac{dy}{dx}+y\,sinx=cosx$$

C

$$x\dfrac{dy}{dx}+y^2\,cosx=5x$$

D

$$\left(\dfrac{dy}{dx}\right)^2=5x$$

Option B is Correct

# Solution to Linear Differential Equation

• A first order linear differential equation is of the form

$$\dfrac{dy}{dx}+y\,P(x)=Q(x)$$

Where $$P(x)$$ and $$Q(x)$$ are functions of $$x$$ and are continuous.

• Linear equations occur frequently in various sciences and therefore the solution to these equation is very important.
• Note that these are not variable separable and will be solved by multiplying both sides by a factor called integrating factor.
• $$\dfrac{dy}{dx}+y\,P(x)=Q(x)$$
• Multiply both sides by the factor

$$I(x)=e^{\int P(x)dx}$$ called the Integrating factor.

• Equation becomes

$$e^{\int P(x)dx}×\dfrac{dy}{dx}+y\,P(x)\,e^{\int P(x)dx}=Q(x)\,e^{\int P(x)dx}$$

$$\dfrac{d}{dx}\left(y\,e^{\int P(x)dx}\right)=Q(x)e^{\int P(x)dx}$$

Now integrate both sides

$$\displaystyle y\,e^{\int P(x)dx}=\int Q(x)e^{\int P(x)dx}$$

If we say $$e^{\int P(x)dx}=$$ (I.F. in short)

$$\displaystyle y×I.F.^{\nearrow^\text{is the general solution}}=\int (Q(x)×I.F.)dx$$

• $$e^{\int P(x)dx}$$ is called the integrating factor because after multiplying by this we are able to integrate both sides.
• If any initial condition is given, we can use this condition in the general solution to find the value of 'c' and then put back the value of 'c' to get particular solution.

#### Find the integrating factor for the linear differential equation $$\dfrac{dy}{dx}+\dfrac{y}{x}=x^2$$

A $$I.F.=sin\,x$$

B $$I.F.=x$$

C $$I.F.=tan^2\,x$$

D $$I.F.=ln \,x$$

×

For the linear differential equation of the form

$$\dfrac{dy}{dx}+y\,P(x)=Q(x)$$ the integrating factor is

$$I.F.=e^{\int P(x)dx}$$

In the given differential equation

$$\dfrac{dy}{dx}+\dfrac{y}{x}=x^2\to P(x)=\dfrac{1}{x}$$ and $$Q(x)=x^2$$

$$\therefore\,I.F.=e^{\int1/x\,dx}=e^{ln\,x}$$

$$=x\,(\because\,{e^{ln\,b}}=b)$$.

### Find the integrating factor for the linear differential equation $$\dfrac{dy}{dx}+\dfrac{y}{x}=x^2$$

A

$$I.F.=sin\,x$$

.

B

$$I.F.=x$$

C

$$I.F.=tan^2\,x$$

D

$$I.F.=ln \,x$$

Option B is Correct

# Bernoulli's Differential Equation

• A differential equation of the form

$$\dfrac{dy}{dx}+y\,P(x)=y^n\,Q(x)$$

is called a Bernoulli's differential equation. It is different from the linear equation because there is an extra factor of $$y^n$$ in the R.H.S.

• For  $$n=0$$ in R.H.S. it reduces to linear differential equation.
• For other values of $$n$$ we first reduce it to linear and then solve it.
• $$\dfrac{dy}{dx}+y\,P(x)=y^2\,Q(x)$$ ...(1)

Step 1- Divide both sides by $$y^n$$

$$\Rightarrow\,$$ ...(1) becomes

$$y^{-n}\dfrac{dy}{dx}+y^{1-n}\,P(x)=Q(x)$$

Step 2- Put $$y^{1-n}=v$$

$$\Rightarrow\,(1-n)\dfrac{dy}{dx}×y^{-n}=\dfrac{dv}{dx}$$

$$\Rightarrow\,y^{-n}\dfrac{dy}{dx}=\dfrac{1}{1-n}\,\dfrac{dv}{dx}$$

$$\therefore$$ Equation ...(1) becomes

$$\dfrac{1}{1-n}\,\dfrac{dv}{dx}+v\,P(x)=Q(x)$$

$$\Rightarrow\,\dfrac{dv}{dx}+v(1-n)\,P(x)=(1-n)\,Q(x)$$...(2)

Step 3- The differential equation (2) obtained is a linear one.

solve $$v$$ in terms of $$x$$ and the put back $$v=y^{1-n}$$

#### Which of the following is a Bernoulli's differential equation?

A $$\dfrac{dy}{dx}+y^2x^2=xsin\,y$$

B $$\dfrac{dy}{dx}+y\,sinx=y^2cosx$$

C $$x\dfrac{dy}{dx}=2ln\,y-y^3$$

D $$\dfrac{dy}{dx}+(siny)x^2=ln\,xy$$

×

A differential equation of the form

$$\dfrac{dy}{dx}+y\,P(x)=y^n\,\,Q(x)$$ is called a Bernoulli differential equation

Only option (b)'s differential equation is in this form with $$n=2$$$$P(x)=sinx$$ and $$Q(x)=cosx$$.

$$\therefore$$ option 'B' is correct.

### Which of the following is a Bernoulli's differential equation?

A

$$\dfrac{dy}{dx}+y^2x^2=xsin\,y$$

.

B

$$\dfrac{dy}{dx}+y\,sinx=y^2cosx$$

C

$$x\dfrac{dy}{dx}=2ln\,y-y^3$$

D

$$\dfrac{dy}{dx}+(siny)x^2=ln\,xy$$

Option B is Correct

#### Solve the Bernoulli differential equation $$\dfrac{dy}{dx}+2xy=2x^3y^3$$

A $$\dfrac{1}{y^2}=x+sinx$$

B $$y^2=ce^x+lnx$$

C $$\dfrac{1}{y^2}=x^2+\dfrac{1}{2}+ce^{2x^2}$$

D $$y^2=\dfrac{1}{5x}+x^2$$

×

To solve Bernoulli equation of the form

$$\dfrac{dy}{dx}+y\,P(x)=y^nQ(x)$$ we divide both sides by $$y^n$$ and then put $$y^{1-n}=v$$ to reduce it to linear differential equation.

In this case the differential equation is

$$\dfrac{dy}{dx}+2xy=2x^3y^3$$ ...(1)

(1) Divide both sides by $$y^3$$

$$\therefore$$ equation (1) because $$\dfrac{1}{y^3}\,\dfrac{dy}{dx}+\dfrac{2x}{y^2}=2x^3$$

$${y^{-3}}\,\dfrac{dy}{dx}+{2x}{y^{-2}}=2x^3$$.....(2)

Put $$y^{-2}=v$$

$$\Rightarrow\,-2y^{-3}\dfrac{dy}{dx}=\dfrac{dv}{dx}$$

$$\therefore$$ equation (2) becomes

$$\dfrac{-1}{2}\dfrac{dv}{dx}+2vx=2x^3$$

$$\Rightarrow\,\dfrac{dv}{dx}-4vx=-4x^3$$ (which is linear)

$$I.F.=e^{\int P(x)dx}$$

$$e^{\int -4x}$$

$$e^{-2x^2}$$

$$\therefore\,$$ General solution is

$$ve^{-2x^2}=\int -4x^3e^{-2x^2}dx$$

on R.H.S. put $$-2x^2=t$$

$$-4x\,dx=dt$$

$$\displaystyle ve^{-2x^2}=\int -\dfrac{t}{2}e^tdt$$

$$ve^{-2x^2}=-\dfrac{1}{2}\left[te^t-e^t\right]+c$$ (Using integration by parts)

$$\Rightarrow\,ve^{-2x^2}=\dfrac{-1}{2}\left[-2x^2e^{-2x^2}-e^{-2x^2}\right]+c$$

$$\Rightarrow\,v=x^2+\dfrac{1}{2}+ce^{2x^2}$$

Put back $$v=\dfrac{1}{y^2}$$

$$\Rightarrow\,\dfrac{1}{y^2}=x^2+\dfrac{1}{2}+ce^{2x^2}$$ (required general solution)

### Solve the Bernoulli differential equation $$\dfrac{dy}{dx}+2xy=2x^3y^3$$

A

$$\dfrac{1}{y^2}=x+sinx$$

.

B

$$y^2=ce^x+lnx$$

C

$$\dfrac{1}{y^2}=x^2+\dfrac{1}{2}+ce^{2x^2}$$

D

$$y^2=\dfrac{1}{5x}+x^2$$

Option C is Correct