Class 12 Commerce Applied Mathematics - Differential Equations

If $y=Ae^{mx}+Be^{nx}$ , show that $\dfrac{d^2y}{dx^2}-(m+n)\dfrac{dy}{dx}+mny=0$ .

Given

$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -(m+n)\dfrac { dy }{ dx } +mny=0$

$Solution:y=A{ e }^{ mx }+B{ e }^{ nx }$

$\dfrac { dy }{ dx } =Am{ e }^{ mx }+Bn{ e }^{ nx }$

$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =A{ m }^{ 2 }{ e }^{ mx }+B{ n }^{ 2 }{ e }^{ nx }$

Substituting the values in above equation we get

$A{ m }^{ 2 }{ e }^{ mx }+B{ n }^{ 2 }{ e }^{ nx }-(m+n)(Am{ e }^{ mx }+Bn{ e }^{ nx })+mn(A{ e }^{ mx }+B{ e }^{ nx })$

$=0$

Solve: $\left( {1 + {y^2}} \right)dx + \left( {x - {e^{ - {{\tan }^{ - 1}}y}}} \right)dy = 0,y\left( 0 \right) = 0$

$(1+y^2)dx+(x-e^{-\tan^{-1}y})dy=0$

$\Rightarrow \dfrac{dx}{dy}+\dfrac{x}{1+y^2}=\dfrac{e^{-\tan^{-1}y}}{1+y^2}$

Integrating factor

$\displaystyle e^{\displaystyle \int\dfrac{1}{1+y^2}dy}=e^{\tan^{-1}y}$

So,

$\displaystyle xe^{\tan^{-1}y}=\int e^{\tan^{-1}y}.\dfrac{e^{-\tan^{-1}y}}{1+y^2}dy=\int \dfrac{1}{1+y^2}dy=\tan^{-1}y+c$

$\therefore x e^{\tan^{-1}y}=\tan^{-1}y+c$

$\Rightarrow x=\tan^{-1}y.e^{-\tan^{-1}y}+ c e^{-\tan^{-1}y}$

Again

$y(0)=0$

$\Rightarrow 0=0.e^0+c.e^0$

$\Rightarrow \boxed{c=0}$

$\therefore \boxed{x=\tan^{-1}y e^{-\tan^{-1}y}}$

Integrating factor of $\cfrac{xdy}{dx}-y={x}^{4}-3x$ is:

$\dfrac{dy}{dx}+\left ( \dfrac{-1}{x} \right )y=x^{3}-3$

intergrating factor

$= e^{\int \left ( \dfrac{-1}{x} \right )dx}$

$=e^{-lnx}$

$=e^{ln \, 1/x}$

$=\dfrac{1}{x}$

The volume of a sphere is increasing at the rate of 3 cubic centimetre per second. Find the rate of its surface area, when the radius is 2cm.

$V=\dfrac{4}{3}\pi r^3$

differentiating on both sides, we get,

$\dfrac{dV}{dt}=4\pi r^2 \dfrac{dr}{dt}$

$\Rightarrow \dfrac{dr}{dt}=\dfrac{1}{4\pi r^2}\dfrac{dV}{dt}$

$r=2cm,\dfrac{dV}{dt}=3cm^2/sec$

$\therefore \dfrac{dr}{dt}=\dfrac{1}{4\pi 2^2}(3)=\dfrac{3}{16\pi}cm/sec$

$S=4\pi r^2$

$\dfrac {dS}{dt}=8 \pi r \dfrac{dr}{dt}$

$=8 \pi (2)\times \dfrac{3}{16\pi}cm/sec$

$\therefore \dfrac {dS}{dt}=3cm^2/sec$

The differential equation of parabola with axis parallel to y-axis and length of latus rectum as 4a is of order:

Let

$\left( u,v \right)$ be the vertex of the parabola.
The equation of required parabolas are of the form

${ \left( x-u \right) }^{ 2 }=4a\left( y-v \right)$
So there are two unknown constant.

Find the integrating factor of the differential equation $\left(\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}-\dfrac{y}{\sqrt{x}}\right)\dfrac{dx}{dy}=1$ .

On rearranging the equation, we get:

$(\dfrac { { e }^{ -2\sqrt { x } } }{ \sqrt { x } } -\dfrac { y }{ \sqrt { x } } )\dfrac { dx }{ dy } =1\\ \Rightarrow \dfrac { dy }{ dx } +\dfrac { y }{ \sqrt { x } } =\dfrac { { e }^{ -2\sqrt { x } } }{ \sqrt { x } }$

which is a first order linear non homogeneous differential equation of the form

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

thus the integrating factor is:

$I.F.={ e }^{ \int { P(x)dx } }={ e }^{ \int { \dfrac { dx }{ \sqrt { x } } } }={ e }^{ 2\sqrt { x } }$

The rate at which the population of a city increases at any time is proportional to the population at that time. If there are $1,30,000$ people in the city in $1960$ and $1,60,000$ in $1990$ , what approximate population may be anticipated in $2020$ ?
$\left[ \log _{ e }{ \cfrac { 16 }{ 13 } } =0.2070,{ e }^{ 0.42 }=1.52 \right]$

Given,

$\cfrac { dx }{ dt } \propto x\Rightarrow \cfrac { dx }{ dt } =kx$

$\therefore \displaystyle \int { \cfrac { dx }{ x } } =\int { k } dt\quad$

$\log { x } =kt+C\Rightarrow x=C{ e }^{ kt }\quad$
At

$t=1960,x=1,30,000$

$\Rightarrow 1,30,000=C{ e }^{ 1960k }$

$\therefore C=\cfrac { 1,30,000 }{ { e }^{ 1960k } }$ ....

$(1)$
At

$t=1990,x=1,60,000$

$\Rightarrow 1,60,000=C{ e }^{ 1960k }$

$=\cfrac { 1,30,000 }{ { e }^{ 1960k } } { e }^{ 1990k }$ From

$(1)$

$=1,30,000{ e }^{ 30k }$

$\therefore { e }^{ 30k }=\cfrac { 1,60,000 }{ 1,30,000 } =\cfrac { 16 }{ 13 }$ ...

$(2)$

$\therefore 30k=\log _{ e }{ \left( \cfrac { 16 }{ 13 } \right) } =0.2070;k=0.007$

$\therefore$

$p=1,30,000$ ;

${e}^{0.007}$ is the population at any time

$t$
In the year

$2020$ , (i.e.,) when

$t=60$

$x=1,30,000\times { e }^{ 0.42 }=1,30,000\times 1.52=1,97,600\quad$
So, the anticipated population is

$2020$ will be

$1,97,600$ .

If the population of a country doubles in $60$ years, in how many years will it be triple under the assumption that the rate of increase is proportional to the number of inhabitants?
[Given: $\log 2 = 0.6912$ and $\log 3 = 1.0986$ .]

Let

$P$ be the number of population

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

$\displaystyle\int\cfrac{dP}P = \int kdt$

$\Rightarrow \ln P = kt + C$ ……….. (Genral Solution)

When

$t=0, P = P_0$

$\ln P_0 = k(0) + C$

$\Rightarrow C = \ln P_0$

When

$t = 60$ years,

$P = 2P_0$

$\ln 2P_0 = k(60) + \ln P_0$

$\Rightarrow k = 0.01152$

When

$t = ?, P = 3P_0$

$\ln 3P_0 = 0.01152(t) + \ln P_0$

$\Rightarrow t = 95$ Years.

The solution of $\dfrac{{dy}}{{dx}} = \dfrac{{x\log {x^2} + x}}{{\sin y + y\cos y}}$ is

We have,

$\dfrac{dy}{dx}=\dfrac{x\log {{x}^{2}}+x}{\sin y+y\cos y}$

$\left( \sin y+y\cos y \right)dy=\left( x\log {{x}^{2}}+x \right)dx$

On taking integral both sides, we get

$\int{\left( \sin y+y\cos y \right)dy}=\int{\left( x\log {{x}^{2}}+x \right)dx}$

$-\cos y+\left[ y\left( \sin y \right)-\int{1.\sin ydy} \right]=\log {{x}^{2}}\left( \dfrac{{{x}^{2}}}{2} \right)-\int{\dfrac{1}{{{x}^{2}}}\times \left( 2x \right)\times \left( \dfrac{{{x}^{2}}}{2} \right)dx}+\dfrac{{{x}^{2}}}{2}+C$

$-\cos y+\left[ y\left( \sin y \right)-\left( -\cos y \right) \right]=\log {{x}^{2}}\left( \dfrac{{{x}^{2}}}{2} \right)-\int{xdx}+\dfrac{{{x}^{2}}}{2}+C$

$-\cos y+\left[ y\left( \sin y \right)+\cos y \right]=\log {{x}^{2}}\left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{2}}}{2}+C$

$y\sin y=\dfrac{{{x}^{2}}}{2}\log {{x}^{2}}+C$

Hence, this is the answer.

Solve $2y\,\,{e^{\dfrac{x}{y}}}\,\,dx\, + \,\left( {y - 2x\,{e^{\dfrac{x}{y}}}} \right)\,\,dy = 0$

We have,

$2y{{e}^{\dfrac{x}{y}}}dx+\left( y-2x{{e}^{\dfrac{x}{y}}} \right)dy=0$

$\dfrac{dx}{dy}=-\dfrac{\left( y-2x{{e}^{\dfrac{x}{y}}} \right)}{2y{{e}^{\dfrac{x}{y}}}}$ ……. (1)

Put $x=vy$

On differentiating w.r.t $y$ , we get

$\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$

From equation (1), we get

$v+y\dfrac{dv}{dy}=-\dfrac{\left( y-2vy{{e}^{v}} \right)}{2y{{e}^{v}}}$

$y\dfrac{dv}{dy}=-\dfrac{\left( y-2vy{{e}^{v}} \right)}{2y{{e}^{v}}}-v$

$y\dfrac{dv}{dy}=-\dfrac{\left( 1-2v{{e}^{v}} \right)}{2{{e}^{v}}}-v$

$y\dfrac{dv}{dy}=\dfrac{-1+2v{{e}^{v}}-2{{e}^{v}}v}{2{{e}^{v}}}$

$y\dfrac{dv}{dy}=\dfrac{-1}{2{{e}^{v}}}$

$2{{e}^{v}}dv=-\dfrac{dy}{y}$

On integration both sides, we get

$\int{2{{e}^{v}}dv}=\int{-\dfrac{dy}{y}}$

$2{{e}^{v}}=-\ln y+C$

$2{{e}^{\dfrac{x}{y}}}+\ln y=C$

Hence, this is the answer.

Solve $\dfrac{dy}{dx}=xcosx$

We have,

$\dfrac{dy}{dx}=x\cos x$

$dy=x\cos xdx$ …….. (1)

On taking integral both sides, we get

$\int{dy}=\int{x\cos xdx}$

$y=x\sin x-\int{1\sin x}dx$

$y=x\sin x-\int{\sin x}dx$

$y=x\sin x-\left( -\cos x \right)+C$

$y=x\sin x+\cos x+C$

Hence, this is the answer.

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was $200000$ in $1990$ and $250000$ in $2000$ , what will be the population in $2010$ ?

Let P be the population at any time t. It is given that

$\dfrac{dP}{dt}\infty P$

$\Rightarrow \dfrac{dP}{dt}=\lambda P$ , where

$\lambda$ is a constant of proportionally

$\Rightarrow \dfrac{dP}{P}=\lambda dt$

$\Rightarrow log P=\lambda t+log C$ ……(i)

Initially i.e., at

$t=1990$ ,

$P=200000$ and at

$t=2000$ ,

$P=250000$

$\therefore \log 200000=1990\lambda + \log C$ ……..(i) and,

$log 250000=2000\lambda +\log C$ ……..(ii)

$\Rightarrow \log 200000-\log 250000=10\lambda$ [On subtracting (ii) from (i)]

$\Rightarrow \lambda =\dfrac{1}{10}log\left(\dfrac{4}{5}\right)$

Putting

$\lambda =\dfrac{1}{10}\log\left(\dfrac{4}{5}\right)$ in (i), we get

$\log \ 200000=1990\times \dfrac{1}{10}log\left(\dfrac{4}{5}\right)+\log C$

$\Rightarrow \log \ 200000=199 \log \dfrac{4}{5}+\log C$

$\Rightarrow log C=\log 200000-199 \log\dfrac{4}{5}$

Putting

$\lambda =\dfrac{1}{10}log \dfrac{4}{5}, log C=\log 200000-199 \ \log\dfrac{4}{5}$ and

$t=2010$ in (i), we get

$log P=\left(\dfrac{1}{10} \log\dfrac{4}{5}\right)2010+\log 200000-199 \log \dfrac{4}{5}$

$\Rightarrow \log P=\log\left(\dfrac{4}{5}\right)^{201}+\log \left(200000\times \left(\dfrac{5}{4}\right)^{199}\right)$

$\Rightarrow P=\left(\dfrac{4}{5}\right)^{201}\times 200000\times \left(\dfrac{5}{4}\right)^{199}=\left(\dfrac{5}{4}\right)^2\times 200000=\dfrac{25 }{16}\times 200000=312500$ .

Represent the following families of curves by forming the corresponding differential equation.(a, b being parameters).
$y^2=4ax$ .

Consider,

$y^2=4ax$

By differentiating we get,

$2y\dfrac {dy}{dx}=4a$

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

$y-2x\dfrac{dy}{dx}=0$ .

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was $200000$ in $1990$ and $250000$ in $2000$ , what will be the population in $2010$ ?

Let P be the population at any time t. It is given that

$\dfrac{dP}{dt}\infty P$

$\Rightarrow \dfrac{dP}{dt}=\lambda P$ , where

$\lambda$ is a constant of proportionally

$\Rightarrow \dfrac{dP}{P}=\lambda dt$

$\Rightarrow log P=\lambda t+log C$ ……(i)

Initially i.e., at

$t=1990$ ,

$P=200000$ and at

$t=2000$ ,

$P=250000$

$\therefore log 200000=1990\lambda +log C$ ……..(i) and,

$log 250000=2000\lambda +log C$ ……..(ii)

$\Rightarrow log 200000-log 250000=10\lambda$ [On subtracting (ii) from (i)]

$\Rightarrow \lambda =\dfrac{1}{10}log\left(\dfrac{4}{5}\right)$

Putting

$\lambda =\dfrac{1}{10}log\left(\dfrac{4}{5}\right)$ in (i), we get

$log 200000=1990\times \dfrac{1}{10}log\left(\dfrac{4}{5}\right)+log C$

$\Rightarrow log 200000=199 log \dfrac{4}{5}+log C$

$\Rightarrow log C=log 200000-199 log\dfrac{4}{5}$

Putting

$\lambda =\dfrac{1}{10}log \dfrac{4}{5}, log C=log 200000-199 \ log\dfrac{4}{5}$

and

$t=2010$ in (i), we get

$log P=\left(\dfrac{1}{10} log\dfrac{4}{5}\right)2010+log 200000-199\ log \dfrac{4}{5}$

$\Rightarrow log P=log\left(\dfrac{4}{5}\right)^{201}+log\left(200000\times \left(\dfrac{5}{4}\right)^{199}\right)$

$\Rightarrow P=\left(\dfrac{4}{5}\right)^{201}\times 200000\times \left(\dfrac{5}{4}\right)^{199}=\left(\dfrac{5}{4}\right)^2\times 200000=\dfrac{25}{16}\times 200000=312500$ .

$\therefore$ The population in

20102010 will be 312500.

Let $\mathrm{f}:\mathrm{R}\rightarrow \mathrm{R}$ be a continuous function which satisfies $\displaystyle \mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}}\mathrm{f}(t)dt$ Then the value of $\mathrm {f(ln5)}$ is

$f(x)=\int _{0}^{x}{f(t) }dt$
Substitute

$x=0$ , then integral of

$f(t)$ from

$0$ to

$0$ will become

$0$ .

$\Rightarrow f(0)=0$
Now, differentiate both side of given integral

${ f }^{ ' }(x)=f(x) , x>0$

$\Rightarrow f(x)=k{ e }^{ x } , x>0$

$\because f(0)=0$ and

$f(x)$ is continuous

$\Rightarrow f(x)=0$

$x>0$

$\therefore f(ln5)=0$

A function $y = f(x)$ satisfying $f"(x) = x^{-3/2}, f'(4) = 2$ and $f(0) = 0$ is $-\lambda \sqrt{x}+3x$ , then find the value of $\lambda$

$\dfrac{d^{2}y}{dx^{2}} = x^{-3/2}$

$\dfrac{dy}{dx} = \dfrac{x^{-1/2}}{-1/2} + c$

$\because \dfrac{dy}{dx} \ at\ x= 4 = -1+c \Rightarrow 2 = c-1 \Rightarrow c= 3$

$\dfrac{dy}{dx} = -\dfrac{2}{\sqrt{x}} + 3$

$y = -2 \dfrac{x^{1/2}}{\dfrac{1}{2}} + 3x+d$

$\because y \ at\ x= 0 =d \Rightarrow 0 = d$

$y = -4 \sqrt{x} + 3x$

The largest value of $\dfrac{2}{\pi}c$ such that there exists a differentiable function $h(x)$ for $-c <x <c$ that is a solution of $y'=1+y^{2}$ with $h(0) = 0$ is:

${ y }^{' }=1+{ y }^{ 2 }\Rightarrow \cfrac { dy }{ 1+{ y }^{ 2 } } =dx\Rightarrow \tan ^{ -1 }{ y } =x+c\\ y\left( 0 \right) =0\Rightarrow c=0\Rightarrow \tan ^{ -1 }{ y } =x$
Since the range of

$\tan ^{ -1 }{ y }$ is

$\left( -\cfrac { \pi }{ 2 } ,\cfrac { \pi }{ 2 } \right)$
The largest value of

$c$ such that

$y=h\left( x \right)$ is differentiable is

$\cfrac { \pi }{ 2 }$

A tank consists of 50 liters of fresh water. Two liters of brine each litre containing 5 gms of dissolved salt are running into tank per minute ; the mixture is kept uniform by stirring , and runs out at the rate of one litre per minute.If $m$ grams of salt are present in the tank after t minute. expression of $m$ in terms of $t$ be $\quad m\, =\, kt \left ( r+\displaystyle \frac {2p+t}{p+t} \right ) gm$ . Find $(p/k)+r$ ?

Input rate = 10 gm/min
After time t volume of tank = 50 + (2-1)t
Concentration of salt in time

$t\, =\, \displaystyle \frac {m}{50+t}\, gm/lit$
Output rate

$=\, \displaystyle \frac {m}{50+t}.1\, gm/min$

$\Rightarrow\displaystyle \frac {dm}{dt}\, =\, \displaystyle \frac {-m}{50+t}\, +\, 10$

$\Rightarrow m(50+t)\, =\, \int 10(50+t)\, dt$

$\Rightarrow m(50+t)\, =\, 500\, t\, +\, 5t^2\, +\, c$

$\Rightarrow (\because \, at\, t\, =\, 0,\, m\, =\, 0\, \Rightarrow\, c\, =\, 0)$

$\Rightarrow m(50+t)\, =\, 5\, (100\, t+t^2)$

$\Rightarrow m\, =\, 5t \left ( \displaystyle \frac {100+t}{50+t} \right ) gm$

Let $(x^2 + 1) \displaystyle \frac{d^2 y}{dx^2} = 2x \frac{dy}{dx},$ where $y'(0) = 3$ & $y(0) = 1$ , then $y(1)$ is equal to:

$\displaystyle \frac{y''}{y'} = \frac{2x}{x^2 + 1}$
On integrating, we get

$\ln y' = \ln (x^2 + 1) + \ln c$

$\Rightarrow y' = c(x^2 + 1)$

$y' (0) = 3$

$\Rightarrow c = 3$

$y' = 3(x^2 + 1)$

$\Rightarrow y = x^3 + 3x + 1$

$\Rightarrow y (1) = 5$

Find the particular solution of the following differential equation;
$(x + 1)\dfrac {dy}{x} = 2e^{-y} - 1; y = 0$ when $x = 0$

Given

$(x+1)\dfrac{dy}{dx}=2e^{-y}-1$

$\Rightarrow \dfrac{dy}{2e^{-y}-1}=\dfrac{dx}{x+1}$ \dfrac

Integrating both the sides, we get

$\displaystyle\int \dfrac{dy}{2e^{-y}-1}=\displaystyle\int \frac{dx}{x+1}$ \dfrac

$\Rightarrow \dfrac{e^y dy}{2-e^y}=\log \left | x+1 \right |+c$

$\Rightarrow -\displaystyle\int\dfrac{dz}{z}=\log \left | x+1 \right |+c$

$\Rightarrow -\log z=\log \left | x+1 \right |+c$

$\Rightarrow -\log \left | 2-e^y \right |=\log \left | x+1 \right |+c$

$\Rightarrow \log \left | x+1 \right |+\log \left | 2-e^y \right |=\log k$

$(x+1)(2-e^y)=k$

Putting x = 0, y = 0, we get

$1.(2-e^0)=k\Rightarrow k = 1$

Therefore, required particular solution is,

$(x+1)(2-e^y)=1$

Find the particular solution of the following differential equation:
$(x+1)\dfrac{dy}{dx} = 2e^{-y} - 1$ , given that $y = 0$ when $x = 0$ .

$(x+1)\dfrac { dy }{ dx } =2{ e }^{ -y }-1\longrightarrow eqn.1\\ \dfrac { dy }{ dx } +\dfrac { 1 }{ 1+x } =\dfrac { 2{ e }^{ -y } }{ 1+x }$

Eqn.1 can be written as:

$\dfrac { 1 }{ 2{ e }^{ -y }-1 } dy=\dfrac { 1 }{ 1+x } dx\\ \Rightarrow \dfrac { 1 }{ \dfrac { 2 }{ { e }^{ y } } -1 } dy=\dfrac { 1 }{ 1+x } dx\\ \Rightarrow \dfrac { -({ -e }^{ y }) }{ 2-{ e }^{ y } } dy=\dfrac { 1 }{ 1+x } dx\\ \Rightarrow (-\log \left| 2-{ e }^{ y } \right| )=\log \left| x+1 \right| +C\\ y=0,\quad x=0\\ \Rightarrow -\log 2=\log 1+C\\ \Rightarrow C=-\log 2$

Answer: $\log (x+1)-\log 2=-\log (2-{ e }^{ -y })\\ \Rightarrow \log (\dfrac { (x+1)(2-{ e }^{ -y }) }{ 2 } )=0$

The temperature $T$ of a cooling object drops at a rate proportional to the difference $(T - S)$ , where $S$ is constant temperature of surrounding medium. If initially $T = 150^{\circ}C$ , find the temperature of the cooling object at any time $'t'$ .

Let

$T$ be the temperature of the cooling object at any time

$t$ .
Given,

$\dfrac {dT}{dt}\propto (T - S) \Rightarrow \dfrac {dT}{dt} = -K(T - S)$ where

$k$ is negative

$\Rightarrow \dfrac {dT}{T - S} =-k. dt$

$\Rightarrow \log (T - S) = -kt + \log c$

$\Rightarrow \log (T - S) = -\log c = kt$

$\Rightarrow \log \dfrac {T - S}{c} = -kt \Rightarrow \log \dfrac {T - S}{c} = e^{-kt}$

$\Rightarrow T - S = c.e^{-kt} \Rightarrow T = S + ce^{-kt}$
When

$t = 0$ and

$T = 150$ ,

$\Rightarrow 150 = S + c \Rightarrow c = 150 - S$

$\therefore$ The temperature of the cooling object at any time

$'t'$ is

$T$ .

$= S + (150 - S) e^{-kt}$

The sum of Rs. $1000$ is compounded continuously, the nominal rate of interest being four percent per annum. In how many years will the amount be twice the original principal? $(\log_e2 = 0.6931)$ .

Let Rs.

$x$ be the principal.
The required differential equation is

$\dfrac{dx}{dt} = \dfrac{4}{100}x$ (or)

$\dfrac{dx}{x} = 0.04 dt$
Integrating both sides, we get

$\displaystyle \int \dfrac{dx}{x} = 0.04 \int dt$

$\Rightarrow \log x$

$= 0.04 t + \log c \Rightarrow x = ce^{0.04t}$ ....... (1)
Given,

$x = 1000$ , when

$t = 0$
Hence (1) becomes

$c = 1000$
when

$x = 2000$

$\Rightarrow 2000$

$= 1000 e^{-0.04t}$

$\Rightarrow 2 = e^{-0.04t}$

$\Rightarrow \log_e 2 = 0.04 t$

$\Rightarrow t = \dfrac{\log_{e}2}{0.04} = \dfrac{0.6931}{0.04} = 17.32 = 17$ years
Thus the number of years required for the amount to be twice the principal is approximately

$17$ years.

A radioactive substance disintegrates at a rate proportional to its mass. When its mass is $10$ mg, the rate of disintegration is $0.051$ mg per day. How long will it take for the mass to be reduced from $10$ mg to $5$ mg? $\left( \log _{ 2 }{ 2 } =0.6931 \right)$ .

Let

$A$ be amount of mass at the time

$t$

$\dfrac { dA }{ dt } \propto A$

$\dfrac { dA }{ dt } =KA$ ........(*)

$\dfrac { dA }{ A } =Kdt$

$\displaystyle\int { \dfrac { dA }{ A } } =\displaystyle\int { Kdt }$

$\Rightarrow \log { A } =Kt+\log { c }$

$\Rightarrow \log { \dfrac { A }{ c } } =Kt$

$\Rightarrow \dfrac { A }{ c } ={ e }^{ Kt }$

$\Rightarrow A=c{ e }^{ Kt }$ ......(1)
when

$t=0$ ,

$A=10$
(1)

$\Rightarrow 10=c{ e }^{ 0 }$

$c=10$

$\therefore$ the solution is

$A=10{ e }^{ Kt }$ .
Given when

$A=10$ ,

$\dfrac { dA }{ dt } =-0.051$ (

$\therefore$ disintegration)

$\Rightarrow -0.051=10k$ (by*)

$\Rightarrow K=-0.0051$

$\therefore A=10{ e }^{ -0.0051t }$
To find

$t$ when

$A=5$

$\therefore A=10{ e }^{ -0.0051t }$
To find

$t$ when

$A=5$

$\Rightarrow 5=10{ e }^{ -0.0051t }$

$\Rightarrow \dfrac { 1 }{ 2 } ={ e }^{ -0.0051t }$

$\Rightarrow 2={ e }^{ -0.0051t }$

$\Rightarrow \log { 2 } =\log { { e }^{ -0.0051t } }$

$\Rightarrow 0.0051t=\log { 2 }$

$\Rightarrow t=\dfrac { \log { 2 } }{ 0.0051 } =\dfrac { 0.6931 }{ 0.0051 } =136$ days.

A body is heated at $110^{\circ}C$ and placed in air at $10^{\circ}C$ . After $1$ hour its temperature is $60^{\circ}C$ . How much additional time is required for it to cool to $35^{\circ}C$ ?

Let

$\theta$ be the temperature of body at time

$t$ . Temperature of air is given to be

$10^{\circ}C$ .
Then, according to Newton's law of cooling, we have

$\dfrac {d\theta}{dt}\propto (\theta - 10^{\circ})$

$\Rightarrow \dfrac {d\theta}{dt} = -k (\theta - 10^{\circ}), k > 0$

$\Rightarrow \dfrac {d\theta}{\theta - 10^{\circ}} = -k dt$
On integrating both sides, we get

$\displaystyle \int \dfrac {d\theta}{\theta - 10^{\circ}} = -k \int dt$

$\ln (\theta - 10^{\circ}) = -kt + c$

$\theta = 10^{\circ} + e^{kt + c}$ .... (i)
when

$t = 0, \theta = 110^{\circ}$
Substituting in the equation, we get

$110^{\circ} = 10^{\circ} + e^{-k(0) + c}$

$e^{c} = 110^{\circ}$
Substituting the above equation in (i), we get

$\theta = 10^{\circ} + 100^{\circ} e^{-kt}$ .... (ii)
Given that

$\theta = 60^{\circ}C$ , when

$t = 1$ , we get

$60^{\circ} = 10^{\circ} + 100^{\circ}e^{-k}$

$\Rightarrow 50^{\circ} = 100^{\circ} e^{-k}$

$\Rightarrow e^{-k} = \dfrac {1}{2}\Rightarrow e^{k} = 2$

$\Rightarrow k = \log 2$
Now, we have to find

$t$ , when

$\theta = 35^{\circ}$
From equation (ii), we have

$\theta = 10^{\circ} + 100^{\circ}e^{-kt}$

$\Rightarrow 35^{\circ} = 10^{\circ} + 100^{\circ} e^{-kt}$

$\Rightarrow e^{-kt} = \dfrac {1}{4}\Rightarrow e^{kt} = 4$

$\Rightarrow kt = \log 4$

$\Rightarrow t = \dfrac {\log 4}{\log 2}$

$\therefore$ Additional time

$= \dfrac {\log 4}{\log 2} - 1 = 1$ hour.

The population of a town grows at the rate of $10\%$ per year. Using differential equation, find how long will it take for the population to grow $4$ times.

Since increase in population speeds up with increase in population and let $x$ be the population at any time $t$

$\therefore \dfrac{dx}{dt} \ \alpha\ x$

$\dfrac{dx}{dt}=r x$

where $r$ is proportionality constant.

Rewriting,

$\dfrac{dx}{x}=rdt$

Integrating,

$ln x=rt +c$

where $c$ is the a constant

Exponentiating on both sides with $e$

$x=e^{rt+c}=Ae^{rt}$ , where $A=e^c$

Let the initial population $x_0$ , then

$x_0=A\dots (1)$

If the population is increased $4$ times in time $t$ , then

$4x_0=Ae^{rt}\dots (2)$

From (1), $x=A$

$4x_0=x_0e^{rt}$

Taking log on both sides

$\ln 4=rt$

$r=0.1$ ...(as rate of increase is given as $10 \%$ )

$t=\dfrac{\ln4}{0.1}=\dfrac{1.38}{0.1}=13.8$ years that is $13$ years and around $10$ months

The rate of growth of population is proportional to the number present. If the population doubled in the last $25$ years and the present population is $1$ lac, when will the city have population $4,00,000$ ?

$\cfrac{dp}{dt}\propto p$

$\implies \cfrac{dp}{dt}=k p$ (where

$k$ is constant)

$\implies \int { \cfrac { dp }{ dt } } =\int { k } dt$

$\implies \ln p=kt+c$ (on integrating both sides)

$t=0,p=(1/2)lac=50,000.$

$t=25,p=(1/2)lac=1,00,000.$

$\therefore$ we get,

$c=\ln(50,000)$ and

$k=\ln(2)/25$

$\therefore$ For

$P=4,00,000$

$\ln p=\cfrac{\ln (2)}{25}t+\ln(50,000)$

$\ln (4,00,000)=\cfrac{\ln (2)}{25}t+\ln(50,000)$

Solving for

$t$ , we get

$t=75$ .

Now as we have assumed

$t=0$ after 25 years. So, city have population of

$4,00,000$ before

$25$ years.

Hence time

$=75-25=50$ years

$(x^{2} + y^{2}) dy = xy dx$ . If $y(1)$ and $y(x_{0})= e$ then find the value of $x_{0}$ .

$(x^{2} + y^{2}) dy = xy dx\Rightarrow \int \dfrac xy dy+\int \dfrac yxdy=\int dx\Rightarrow x\log y+\dfrac{y^2}{2x}=x+c$

Now, at

$x=1 ;y=e$

$x\log y+\dfrac{y^2}{2x}=x+c\Rightarrow x+\dfrac{e^2}{2}=x+c\Rightarrow c=\dfrac{e^2}{2}$

Now at

$x=x_0;y=e$

$x_0\log e+\dfrac{e^2}{2x_0}=x_0+\dfrac{e^2}{2}\Rightarrow \dfrac{e^2}{2x_0}=\dfrac{e^2}{2}\Rightarrow x_0=1$

Differential Equations - Class 12 Commerce Applied Mathematics - Extra Questions

If $y=Ae^{mx}+Be^{nx}$ , show that $\dfrac{d^2y}{dx^2}-(m+n)\dfrac{dy}{dx}+mny=0$ .

Solve: $\left( {1 + {y^2}} \right)dx + \left( {x - {e^{ - {{\tan }^{ - 1}}y}}} \right)dy = 0,y\left( 0 \right) = 0$

Integrating factor of $\cfrac{xdy}{dx}-y={x}^{4}-3x$ is:

The volume of a sphere is increasing at the rate of 3 cubic centimetre per second. Find the rate of its surface area, when the radius is 2cm.

The differential equation of parabola with axis parallel to y-axis and length of latus rectum as 4a is of order:

Find the integrating factor of the differential equation $\left(\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}-\dfrac{y}{\sqrt{x}}\right)\dfrac{dx}{dy}=1$ .

If the population of a country doubles in $60$ years, in how many years will it be triple under the assumption that the rate of increase is proportional to the number of inhabitants?
[Given: $\log 2 = 0.6912$ and $\log 3 = 1.0986$ .]

The solution of $\dfrac{{dy}}{{dx}} = \dfrac{{x\log {x^2} + x}}{{\sin y + y\cos y}}$ is

Solve $2y\,\,{e^{\dfrac{x}{y}}}\,\,dx\, + \,\left( {y - 2x\,{e^{\dfrac{x}{y}}}} \right)\,\,dy = 0$

Solve $\dfrac{dy}{dx}=xcosx$

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was $200000$ in $1990$ and $250000$ in $2000$ , what will be the population in $2010$ ?

Represent the following families of curves by forming the corresponding differential equation.(a, b being parameters).
$y^2=4ax$ .

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was $200000$ in $1990$ and $250000$ in $2000$ , what will be the population in $2010$ ?

Let $\mathrm{f}:\mathrm{R}\rightarrow \mathrm{R}$ be a continuous function which satisfies $\displaystyle \mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}}\mathrm{f}(t)dt$ Then the value of $\mathrm {f(ln5)}$ is

A function $y = f(x)$ satisfying $f"(x) = x^{-3/2}, f'(4) = 2$ and $f(0) = 0$ is $-\lambda \sqrt{x}+3x$ , then find the value of $\lambda$

The largest value of $\dfrac{2}{\pi}c$ such that there exists a differentiable function $h(x)$ for $-c <x <c$ that is a solution of $y'=1+y^{2}$ with $h(0) = 0$ is:

Let $(x^2 + 1) \displaystyle \frac{d^2 y}{dx^2} = 2x \frac{dy}{dx},$ where $y'(0) = 3$ & $y(0) = 1$ , then $y(1)$ is equal to:

Find the particular solution of the following differential equation;
$(x + 1)\dfrac {dy}{x} = 2e^{-y} - 1; y = 0$ when $x = 0$

Find the particular solution of the following differential equation:
$(x+1)\dfrac{dy}{dx} = 2e^{-y} - 1$ , given that $y = 0$ when $x = 0$ .

The temperature $T$ of a cooling object drops at a rate proportional to the difference $(T - S)$ , where $S$ is constant temperature of surrounding medium. If initially $T = 150^{\circ}C$ , find the temperature of the cooling object at any time $'t'$ .

The sum of Rs. $1000$ is compounded continuously, the nominal rate of interest being four percent per annum. In how many years will the amount be twice the original principal? $(\log_e2 = 0.6931)$ .

A radioactive substance disintegrates at a rate proportional to its mass. When its mass is $10$ mg, the rate of disintegration is $0.051$ mg per day. How long will it take for the mass to be reduced from $10$ mg to $5$ mg? $\left( \log _{ 2 }{ 2 } =0.6931 \right)$ .

A body is heated at $110^{\circ}C$ and placed in air at $10^{\circ}C$ . After $1$ hour its temperature is $60^{\circ}C$ . How much additional time is required for it to cool to $35^{\circ}C$ ?

The population of a town grows at the rate of $10\%$ per year. Using differential equation, find how long will it take for the population to grow $4$ times.

The rate of growth of population is proportional to the number present. If the population doubled in the last $25$ years and the present population is $1$ lac, when will the city have population $4,00,000$ ?

$(x^{2} + y^{2}) dy = xy dx$ . If $y(1)$ and $y(x_{0})= e$ then find the value of $x_{0}$ .

Class 12 Commerce Applied Mathematics Extra Questions

Differential Equations - Class 12 Commerce Applied Mathematics - Extra Questions

If y=Aemx+Benxy=Ae^{mx}+Be^{nx}, show that d2ydx2−(m+n)dydx+mny=0\dfrac{d^2y}{dx^2}-(m+n)\dfrac{dy}{dx}+mny=0.

Solve: (1+y2)dx+(x−e−tan−1y)dy=0,y(0)=0\left( {1 + {y^2}} \right)dx + \left( {x - {e^{ - {{\tan }^{ - 1}}y}}} \right)dy = 0,y\left( 0 \right) = 0

Integrating factor of xdydx−y=x4−3x\cfrac{xdy}{dx}-y={x}^{4}-3x is:

The volume of a sphere is increasing at the rate of 3 cubic centimetre per second. Find the rate of its surface area, when the radius is 2cm.

The differential equation of parabola with axis parallel to y-axis and length of latus rectum as 4a is of order:

Find the integrating factor of the differential equation (e−2√x√x−y√x)dxdy=1\left(\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}-\dfrac{y}{\sqrt{x}}\right)\dfrac{dx}{dy}=1.

If the population of a country doubles in 6060 years, in how many years will it be triple under the assumption that the rate of increase is proportional to the number of inhabitants?[Given: log2=0.6912\log 2 = 0.6912 and log3=1.0986\log 3 = 1.0986.]

The solution of dydx=xlogx2+xsiny+ycosy\dfrac{{dy}}{{dx}} = \dfrac{{x\log {x^2} + x}}{{\sin y + y\cos y}} is

Solve 2yexydx+(y−2xexy)dy=02y\,\,{e^{\dfrac{x}{y}}}\,\,dx\, + \,\left( {y - 2x\,{e^{\dfrac{x}{y}}}} \right)\,\,dy = 0

Solve dydx=xcosx\dfrac{dy}{dx}=xcosx

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000200000 in 19901990 and 250000250000 in 20002000, what will be the population in 20102010?

Represent the following families of curves by forming the corresponding differential equation.(a, b being parameters).y2=4axy^2=4ax.

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000200000 in 19901990 and 250000250000 in 20002000, what will be the population in 20102010?

Let f:R→R\mathrm{f}:\mathrm{R}\rightarrow \mathrm{R} be a continuous function which satisfies f(x)=∫x0f(t)dt\displaystyle \mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}}\mathrm{f}(t)dt Then the value of f(ln5)\mathrm {f(ln5)} is

A function y=f(x)y = f(x) satisfying f"(x) = x^{-3/2}, f'(4) = 2 f"(x) = x^{-3/2}, f'(4) = 2 and f(0) = 0 f(0) = 0 is -\lambda \sqrt{x}+3x -\lambda \sqrt{x}+3x, then find the value of \lambda\lambda

The largest value of \dfrac{2}{\pi}c\dfrac{2}{\pi}c such that there exists a differentiable function h(x)h(x) for -c <x <c-c <x <c that is a solution of y'=1+y^{2}y'=1+y^{2} with h(0) = 0h(0) = 0 is:

Let (x^2 + 1) \displaystyle \frac{d^2 y}{dx^2} = 2x \frac{dy}{dx},(x^2 + 1) \displaystyle \frac{d^2 y}{dx^2} = 2x \frac{dy}{dx}, where y'(0) = 3y'(0) = 3 & y(0) = 1y(0) = 1, then y(1)y(1) is equal to:

Find the particular solution of the following differential equation;(x + 1)\dfrac {dy}{x} = 2e^{-y} - 1; y = 0(x + 1)\dfrac {dy}{x} = 2e^{-y} - 1; y = 0 when x = 0x = 0

Find the particular solution of the following differential equation:(x+1)\dfrac{dy}{dx} = 2e^{-y} - 1(x+1)\dfrac{dy}{dx} = 2e^{-y} - 1, given that y = 0y = 0 when x = 0 x = 0.

The temperature TT of a cooling object drops at a rate proportional to the difference (T - S)(T - S), where SS is constant temperature of surrounding medium. If initially T = 150^{\circ}CT = 150^{\circ}C, find the temperature of the cooling object at any time 't''t'.

The sum of Rs. 10001000 is compounded continuously, the nominal rate of interest being four percent per annum. In how many years will the amount be twice the original principal? (\log_e2 = 0.6931)(\log_e2 = 0.6931).

A body is heated at 110^{\circ}C110^{\circ}C and placed in air at 10^{\circ}C10^{\circ}C. After 11 hour its temperature is 60^{\circ}C60^{\circ}C. How much additional time is required for it to cool to 35^{\circ}C35^{\circ}C?

The population of a town grows at the rate of 10\%10\% per year. Using differential equation, find how long will it take for the population to grow 44 times.

The rate of growth of population is proportional to the number present. If the population doubled in the last 2525 years and the present population is 11 lac, when will the city have population 4,00,0004,00,000 ?

(x^{2} + y^{2}) dy = xy dx(x^{2} + y^{2}) dy = xy dx. If y(1)y(1) and y(x_{0})= ey(x_{0})= e then find the value of x_{0}x_{0}.

Class 12 Commerce Applied Mathematics Extra Questions

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If $y=Ae^{mx}+Be^{nx}$ , show that $\dfrac{d^2y}{dx^2}-(m+n)\dfrac{dy}{dx}+mny=0$ .

Solve: $\left( {1 + {y^2}} \right)dx + \left( {x - {e^{ - {{\tan }^{ - 1}}y}}} \right)dy = 0,y\left( 0 \right) = 0$

Integrating factor of $\cfrac{xdy}{dx}-y={x}^{4}-3x$ is:

Find the integrating factor of the differential equation $\left(\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}-\dfrac{y}{\sqrt{x}}\right)\dfrac{dx}{dy}=1$ .

If the population of a country doubles in $60$ years, in how many years will it be triple under the assumption that the rate of increase is proportional to the number of inhabitants?
[Given: $\log 2 = 0.6912$ and $\log 3 = 1.0986$ .]

The solution of $\dfrac{{dy}}{{dx}} = \dfrac{{x\log {x^2} + x}}{{\sin y + y\cos y}}$ is

Solve $2y\,\,{e^{\dfrac{x}{y}}}\,\,dx\, + \,\left( {y - 2x\,{e^{\dfrac{x}{y}}}} \right)\,\,dy = 0$

Solve $\dfrac{dy}{dx}=xcosx$

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was $200000$ in $1990$ and $250000$ in $2000$ , what will be the population in $2010$ ?

Represent the following families of curves by forming the corresponding differential equation.(a, b being parameters).
$y^2=4ax$ .

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was $200000$ in $1990$ and $250000$ in $2000$ , what will be the population in $2010$ ?

Let $\mathrm{f}:\mathrm{R}\rightarrow \mathrm{R}$ be a continuous function which satisfies $\displaystyle \mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}}\mathrm{f}(t)dt$ Then the value of $\mathrm {f(ln5)}$ is

A function $y = f(x)$ satisfying $f"(x) = x^{-3/2}, f'(4) = 2$ and $f(0) = 0$ is $-\lambda \sqrt{x}+3x$ , then find the value of $\lambda$

The largest value of $\dfrac{2}{\pi}c$ such that there exists a differentiable function $h(x)$ for $-c <x <c$ that is a solution of $y'=1+y^{2}$ with $h(0) = 0$ is:

Let $(x^2 + 1) \displaystyle \frac{d^2 y}{dx^2} = 2x \frac{dy}{dx},$ where $y'(0) = 3$ & $y(0) = 1$ , then $y(1)$ is equal to:

Find the particular solution of the following differential equation;
$(x + 1)\dfrac {dy}{x} = 2e^{-y} - 1; y = 0$ when $x = 0$

Find the particular solution of the following differential equation:
$(x+1)\dfrac{dy}{dx} = 2e^{-y} - 1$ , given that $y = 0$ when $x = 0$ .

The temperature $T$ of a cooling object drops at a rate proportional to the difference $(T - S)$ , where $S$ is constant temperature of surrounding medium. If initially $T = 150^{\circ}C$ , find the temperature of the cooling object at any time $'t'$ .

The sum of Rs. $1000$ is compounded continuously, the nominal rate of interest being four percent per annum. In how many years will the amount be twice the original principal? $(\log_e2 = 0.6931)$ .

A body is heated at $110^{\circ}C$ and placed in air at $10^{\circ}C$ . After $1$ hour its temperature is $60^{\circ}C$ . How much additional time is required for it to cool to $35^{\circ}C$ ?

The population of a town grows at the rate of $10\%$ per year. Using differential equation, find how long will it take for the population to grow $4$ times.

The rate of growth of population is proportional to the number present. If the population doubled in the last $25$ years and the present population is $1$ lac, when will the city have population $4,00,000$ ?

$(x^{2} + y^{2}) dy = xy dx$ . If $y(1)$ and $y(x_{0})= e$ then find the value of $x_{0}$ .