Informative line

### Application Of Differential Equation

Learn how to solve logistic differential equation & orthogonal trajectories differential equations. Practice to find the orthogonal trajectory of the family of curve & population of a certain area is modeled by the logistic differential equation.

# Application to Electric Circuit (Differential Equation)

Inductor-Resistor Circuit :

Consider a simple electric circuit as shown.  Let E be the battery which causes the current $$I(t)$$ to flow in the circuit. The current is the function of time.

• By Kirchhoff's rule

$$E-\displaystyle\underbrace {RI}_{\text{potential drop across resistor}}-\underbrace {L\;dI/dt}_{\text{potential drop across inductor}}=0$$

$$\Rightarrow \,L\dfrac{dI}{dt}+RI=E$$

$$\Rightarrow\,\dfrac{dI}{dt}+\dfrac{RI}{L}=\dfrac{E}{L}$$

which is a linear differential equation in which independent variable is $$t$$(time) and dependent variable is $$I$$ (current).

• Solving this will give the current in the circuit as a function of time.
• First we find the integrating factor and then write the solution.
• For equation of the form $$\to \dfrac{dy}{dx}+y\,P(x)=Q(x)$$

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

and solution is  $$y×I.F.=\int (Q×I.F.)dx$$

#### Consider the figure as shown. The value of resistance $$R$$ is $$10\Omega$$, and the inductance $$L$$ is $$5H$$. If battery given a constant voltage of $$70V$$, switch is closed when $$t=0,\,I(0)=0$$ i.e. no current in the beginning find the current in the circuit after 1 sec. $$[e^{-2}=0.1353]$$

A 15 A

B 6.053 A

C 10 A

D 2.31 A

×

By Kirchhoff law the differential equation

$$\dfrac{dI}{dt}+\dfrac{RI}{L}=\dfrac{E}{L}$$ will be followed.

Putting the values given in the problem

$$R=10,\,L=5,\,E=70$$ we get

$$\dfrac{dI}{dt}+\dfrac{10I}{5}=\dfrac{70}{5}$$

$$\Rightarrow\,\dfrac{dI}{dt}+2I=14$$ (which is linear differential equation)

$$I.F.=e^{\int P(t)dt}$$

$$=e^{\int 2dt}$$

$$=e^{2t}$$

$$\therefore$$ Solution is $$I(t)×e^{2t}=\int14e^{2t}dt$$

$$\Rightarrow\,I(t)e^{2t}=\dfrac{14e^{2t}}{2}+c$$

$$\Rightarrow\,I(t)×e^{2t}=7e^{2t}+c$$

$$\Rightarrow\,I(t)=7+ce^{-2t}$$

Use the initial condition $$\to$$ when $$t=0,\,I=0$$

$$\Rightarrow\,0=7+c$$

$$\Rightarrow\,c=-7$$

$$\Rightarrow \,I(t)=7-7e^{-2t}$$

$$\Rightarrow \,I(t)=7(1-e^{-2t})$$

$$\therefore\,I(1)$$= value asked = $$7(1-e^{-2})=7(1-0.1353)=7×0.865$$

$$=6.053\, Amp.$$

### Consider the figure as shown. The value of resistance $$R$$ is $$10\Omega$$, and the inductance $$L$$ is $$5H$$. If battery given a constant voltage of $$70V$$, switch is closed when $$t=0,\,I(0)=0$$ i.e. no current in the beginning find the current in the circuit after 1 sec. $$[e^{-2}=0.1353]$$ A

15 A

.

B

6.053 A

C

10 A

D

2.31 A

Option B is Correct

# Models of Population Growth

## The law of natural growth

The elementary model for growth of population is based on the assumption that population grows at rate proportional to the size of present population.

P = number of individuals in the population (dependent variable).

t = time (the independent variable)

then  $$\dfrac{dP}{dt}=kP$$, where k is the position proportionality constant.

• We say this is the differential equation of population growth. It is a simple variable separable equation.

$$\dfrac{dP}{dt}=kP$$

$$\Rightarrow \dfrac{dP}{P}=k\,dt$$

Integrating both sides we get

$$\displaystyle\int\dfrac{dP}{P}=\int k\,dt$$

$$\Rightarrow\,ln\,P=kt+C$$

$$\Rightarrow\,P(t)=e^{kt+C}$$

$$\Rightarrow\,P(t)=Ce^{kt}$$

If initially i.e. at $$t=0$$, the population is P0 then

$$P_0=Ce^0\Rightarrow\,C=P_0$$

$$\Rightarrow\,P(t)=P_0e^{kt}$$

$$\therefore\,P(t)=P_0e^{kt}$$

#### The population of a certain place is known to increase at the rate proportional to the number of people presently living in that place. If after two years the population has doubled and after three years the population in 20000. Find the population today.  [ $$ln2=0.6931$$ ,  ​$$e^{1.041}= 2.832$$ ]

A 15181

B 7062

C 56

D 25000

×

Apply the natural growth model

$$\dfrac{dP}{dt}=kP$$ whose solution is

$$P=P_0e^{kt}$$ ...(1)

Now when $$t=2,\,P=2P_0\to$$ put in equation ...(1)

$$\therefore\,2P_0=P_0e^{2k}$$

$$\Rightarrow\,e^{2k}=2$$

$$\Rightarrow\,2k=ln2$$

$$\Rightarrow\,k=\dfrac{1}{2}ln2=\dfrac{1}{2}(0.6931)=0.347$$

When $$t=3,\,P=20000\to$$ put in equation ...(1)

$$20000=P_0e^{3×0.347}$$

$$\Rightarrow\,P_0=\dfrac{20000}{e^{3×0.347}}$$

$$\Rightarrow\,P_0=\dfrac{20000}{e^{0.041}}$$

$$\Rightarrow\,P_0=\dfrac{20000}{2.832}=7062$$

### The population of a certain place is known to increase at the rate proportional to the number of people presently living in that place. If after two years the population has doubled and after three years the population in 20000. Find the population today.  [ $$ln2=0.6931$$ ,  ​$$e^{1.041}= 2.832$$ ]

A

15181

.

B

7062

C

56

D

25000

Option B is Correct

#### The population of a certain area is modeled by the differential equation $$\dfrac{dP}{dt}=1.5P\,\left(1-\dfrac{P}{6000}\right)$$ For when values of $$P$$ is the population increasing ?

A $$P<15000$$

B $$P<6000$$

C $$P>20000$$

D $$P>2$$

×

population is increasing when  $$\dfrac{dP}{dt}>0$$

In this case

$$\dfrac{dP}{dt}=1.5P\,\left(1–\dfrac{P}{6000}\right)$$

$$\therefore\,\dfrac{dP}{dt}>0$$

$$\Rightarrow\,1.5P\left(1–\dfrac{P}{6000}\right)>0$$

$$\Rightarrow\,1-\dfrac{P}{6000}>0$$

$$\Rightarrow\,P<6000$$

### The population of a certain area is modeled by the differential equation $$\dfrac{dP}{dt}=1.5P\,\left(1-\dfrac{P}{6000}\right)$$ For when values of $$P$$ is the population increasing ?

A

$$P<15000$$

.

B

$$P<6000$$

C

$$P>20000$$

D

$$P>2$$

Option B is Correct

#### The population of a certain area is modeled by the differential equation $$\dfrac{dP}{dt}=1.5P\,\left(1-\dfrac{P}{6000}\right)$$ For when values of $$P$$ is the population decreasing ?

A $$P<500$$

B $$P>6000$$

C $$P>5000$$

D $$P<20000$$

×

Population is decreasing when  $$\dfrac{dP}{dt}<0$$

In this case

$$\dfrac{dP}{dt}=1.5P\,\left(1–\dfrac{P}{6000}\right)$$

$$\therefore\,\dfrac{dP}{dt}<0$$

$$\Rightarrow\,1.5P\left(1–\dfrac{P}{6000}\right)<0$$

$$\Rightarrow\,1-\dfrac{P}{6000}<0$$

$$\Rightarrow\,P>6000$$

### The population of a certain area is modeled by the differential equation $$\dfrac{dP}{dt}=1.5P\,\left(1-\dfrac{P}{6000}\right)$$ For when values of $$P$$ is the population decreasing ?

A

$$P<500$$

.

B

$$P>6000$$

C

$$P>5000$$

D

$$P<20000$$

Option B is Correct

# Models of Population Growth (Logistics Model)

• Logistic model

In the natural growth model

$$\dfrac{dP}{dt}=kP$$ when P is the population at any time t.

• Initially the population increases exponentially (according to natural growth model) but it levels off eventually because of limited resources. So $$\dfrac{dP}{dt}\cong kP$$ if P is small.
• As P increases and approaches M which is the carrying capacity, (the maximum population the environment is capable of sustaining) the rate of change of population does not follow the natural growth model.
• $$\therefore$$ we have logistics model

$$\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)$$

• If P > M then $$\dfrac{dP}{dt}<0$$ and the population will decrease.
• As long as P < M and $$\dfrac{dP}{dt}>0$$,the population will increase.

• Consider the logistics equation

$$\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)$$

$$\Rightarrow\,\dfrac{dP}{dt}=\dfrac{kP(M-P)}{M}$$

$$\dfrac{M}{P(M-P)}dP=k\,dt$$ (separate the variable)

Integrate both sides

$$\displaystyle\int \dfrac{MdP}{P(M-P)}=kt+C$$

$$\displaystyle\,\Rightarrow\,\int\left(\dfrac{1}{P}+\dfrac{1}{M-P}\right)dP=kt+C$$ (partial fraction method)

$$\Rightarrow\,lnP+ln|M-P|×-1=kt+C$$

$$\Rightarrow\,ln\left|\dfrac{P}{M-P}\right|=kt+C$$

$$\Rightarrow\,ln\left|\dfrac{M-P}{P}\right|=-kt-C$$

$$\Rightarrow\,\left|\dfrac{M-P}{P}\right|=e^{-kt-C}$$

$$\Rightarrow\,\left|\dfrac{M-P}{P}\right|=Ae^{-kt}(e^{-C}=A)$$

$$\Rightarrow\,\dfrac{M}{P}-1=Ae^{-kt}$$

$$\Rightarrow\,\dfrac{M}{P}=1+Ae^{-kt}$$

$$\Rightarrow \,P=\dfrac{M}{1+Ae^{-kt}}$$

Now when $$t=0$$ (Initially), $$P=P_0$$ So.

$$P_0=\dfrac{M}{1+A}$$

$$\Rightarrow\,A=\dfrac{M-P_0}{P_0}$$

$$\therefore$$ solution to logistic model is

$$P(t)=\dfrac{M}{1+Ae^{-kt}}$$  when $$A=\dfrac{M-P_0}{P_0}$$

Graph showing various values of $${P_0}$$  and population eventually reaching M.  #### Suppose population at a certain place develops according to logistic model. $$\dfrac{dP}{dt}=0.07P-0.0035P^2$$ where t is measured in weeks. What is the carrying capacity M

A $$M=2$$

B $$M=20$$

C $$M=500$$

D $$M=36$$

×

In logistic model, the standard form of differential equation is

$$\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)$$ ...(1)

Put the given equation in this form.

$$\therefore\,\dfrac{dP}{dt}=0.07P-0.0035P^2$$

$$\Rightarrow\,\dfrac{dP}{dt}=0.07P(1-0.05P)$$

$$\Rightarrow\,\dfrac{dP}{dt}=0.07P\left(1-\dfrac{P}{20}\right)$$

$$\therefore$$ comparing with (1)

$$k=0.07,\,M=20$$

### Suppose population at a certain place develops according to logistic model. $$\dfrac{dP}{dt}=0.07P-0.0035P^2$$ where t is measured in weeks. What is the carrying capacity M

A

$$M=2$$

.

B

$$M=20$$

C

$$M=500$$

D

$$M=36$$

Option B is Correct

#### Find the population $$P(50)$$ for the logistics model differential equation. $$\dfrac{dP}{dt}=0.05P(1-\dfrac{P}{10000})$$, $$P(0)=1000$$ where  $$t$$ is in weeks i.e. population after 50 weeks. [ ​​$$e^{-2.5}=0.082$$ ]

A 1250

B 5750

C 15120

D 7805

×

The solution to the logistics model

$$\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)$$

is given by $$P(t)=\dfrac{M}{1+Ae^{-kt}}$$ where  $$A=\dfrac{M-P_0}{P_0}$$

In this case $$M=10000,\,P_0=1000$$

$$\therefore\,A=\dfrac{10000–1000}{1000}=9$$ and  $$k=0.05$$

$$\therefore$$ $$P(t)=\dfrac{10000}{1+9e^{-kt}}$$

put   $$k=0.05$$

$$\Rightarrow \,P(t)=\dfrac{10000}{1+9e^{-0.05t}}$$

$$\therefore\,P(50)=\dfrac{10000}{1+9e^{-0.05×50}}$$

$$=\dfrac{10000}{1+9e^{-2.5}}$$

$$\because\,e^{-2.5}=0.082$$

$$=\dfrac{10000}{1.739}=5750$$

### Find the population $$P(50)$$ for the logistics model differential equation. $$\dfrac{dP}{dt}=0.05P(1-\dfrac{P}{10000})$$, $$P(0)=1000$$ where  $$t$$ is in weeks i.e. population after 50 weeks. [ ​​$$e^{-2.5}=0.082$$ ]

A

1250

.

B

5750

C

15120

D

7805

Option B is Correct

# Orthogonal Trajectories

The orthogonal trajectory of family of curves is a curve that intersects each curve of the family orthogonally that is at right angles.

e.g  $$y=2x$$ line is an orthogonal trajectory to the family of curves  $$x^2+y^2=r^2$$  • In general, a family of curves is called orthogonal trajectories of another family of curves if each member of one family is orthogonal to every member of the other.  • Each line of the family $$y=mx$$ is orthogonal to each member of the circle family $$x^2+y^2=r^2$$.

i.e. the two families of curves are orthogonal trajectories of each other.

• To find orthogonal trajectories of a family of curves
1. First, find the differential equation of the given family of curves by eliminating 'c'.
2. In this differential equation replace $$\dfrac{dy}{dx}$$ by  $$\dfrac{–1}{\dfrac{dy}{dx}}$$.
3. Solve the new differential equation obtained, this will be the orthogonal trajectory.

#### Find the orthogonal trajectory of the family of curve $$y^2=4(x–a)$$ where 'a' is a parameter.

A $$y^2=4Cx+7$$

B $$y=Cx^2$$

C $$y=C\,sinx+cosx$$

D $$y=Ce^{-x/2}$$

×

Find the differential equation of family of curves which is given  $$y^2=4(x-a)$$

$$\Rightarrow\,2yy'=4$$

$$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{2}{y}$$

Replace $$\dfrac{dy}{dx}$$ by $$\dfrac{\dfrac{–1}{dy}}{dx}$$

$$\Rightarrow\,\dfrac{\dfrac{–1}{dy}}{dx}=\dfrac{2}{y}$$

$$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{-y}{2}$$ (differential equation required family)

Now solve this equation

$$\dfrac{dy}{dx}=\dfrac{-y}{2}$$

$$\Rightarrow\,\dfrac{dy}{y}=\dfrac{–1}{2}dx$$ (variable separable)

$$\Rightarrow\,\ln y=\dfrac{–1}{2}x+C$$

$$\Rightarrow\,y=e^{-x/2+C}$$

$$\Rightarrow\,y=e^{\dfrac{-1}{2}{x}}×e^C$$

$$y=Ce^{-x/2}$$

### Find the orthogonal trajectory of the family of curve $$y^2=4(x–a)$$ where 'a' is a parameter.

A

$$y^2=4Cx+7$$

.

B

$$y=Cx^2$$

C

$$y=C\,sinx+cosx$$

D

$$y=Ce^{-x/2}$$

Option D is Correct

# Mixing Problem

A mixing problem involves a tank of fixed capacity (volume) filled with mixed solution of some substance such as salt.

Now a solution of given concentration enters the tank at a fixed rate and the mixture thoroughly stirred leaves at a fixed rate, which is usually different from the entering rate.

If  $$y(t)$$ is the amount of substances in the tank at time 't' then $$y'(t)$$ is the rate at which substance is being added minus the rate at which it is removal.

$$y'(t)$$ = (rate of supply) – (rate of removal)

• This will lead to a differential equation of variable separable category and using initial condition. We can find a particular solution.

#### A tank contains 50 kg of salt dissolved in $$10000 \,\ell$$ of water. Brine which contains 0.05 kg of salt per liter of water enters the tank at the rate of $$25\,l/min.$$ The solution is thoroughly mixed and drain from the tank at some rate. Find the amount of salt in the tank after 20 min. [ $$e^{-1/20}=0.9512$$ ]

A 54.83 kg

B 71.96 kg

C 2 kg

D 78 kg

×

$$y'(t)$$ = rate of supply – rate of removal

Rate of supply of salt = $$\dfrac{0.05kg}{l}×\dfrac{25\,l}{min}$$

$$=1.25\,kg/min$$

Rate of removal of salt = $$\underbrace { \dfrac{y(t)}{10000}}_{\text{concentration at time }'t'}×\dfrac{25\,l}{min}$$

$$\dfrac{dy}{dt}=\dfrac{y(t)}{400}kg/min$$

$$\therefore\,y'(t)=\dfrac{dy}{dt}$$

$$\Rightarrow\,\dfrac{dy}{dt}=1.25–\dfrac{y(t)}{400}$$

$$\Rightarrow\,\dfrac{dy}{dt}=\dfrac{500–y(t)}{400}$$

which is variable separable.

$$\therefore\,\dfrac{dy}{dt}=\dfrac{500–y}{400}$$

$$\Rightarrow\,\dfrac{1}{500–y}dy=\dfrac{1}{400}dt$$

Integrate both side

$$\displaystyle\int\dfrac{1}{500–y}\,dy=\int\dfrac{1}{400}dt$$

$$\Rightarrow\,–ln|500–y|=\dfrac{t}{400}+c$$

Now  $$y(0)=50$$

$$\Rightarrow\,–ln(500–50)=\dfrac{t}{400}+c$$

$$\Rightarrow\,c=-\ln450$$

$$\therefore\,–ln|500–y|=\dfrac{t}{400}–ln(450)$$

$$\Rightarrow\,ln\left[\dfrac{500–y}{450}\right]=\dfrac{–t}{400}$$

$$|500–y|=450e^{–t/400}$$ (both sides one position)

$$\Rightarrow\,y=500–450e^{-t/400}$$

$$\Rightarrow\,y(t)=500–450e^{–t/400}$$

$$\therefore\,$$ The amount of salt after 20 min.

$$=y(20)=500–450×e^{–20/400}$$

$$=500–450×e^{–1/20}$$

$$=500–450×0.9512$$

$$=71.96\,kg$$

### A tank contains 50 kg of salt dissolved in $$10000 \,\ell$$ of water. Brine which contains 0.05 kg of salt per liter of water enters the tank at the rate of $$25\,l/min.$$ The solution is thoroughly mixed and drain from the tank at some rate. Find the amount of salt in the tank after 20 min. [ $$e^{-1/20}=0.9512$$ ]

A

54.83 kg

.

B

71.96 kg

C

2 kg

D

78 kg

Option B is Correct