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Differential Equations

JEE Mathematics · 179 questions · Page 2 of 18 · Click an option or "Show Solution" to reveal answer

Q11
The differential equation whose solution is Ax2+By2=1A{x^2} + B{y^2} = 1 where AA and BB are arbitrary constants is of
A second order and second degree
B first order and second degree
C first order and first degree
D second order and first degree
Correct Answer
Option D
Solution
Ax2+By2=1              ...(1)A{x^2} + B{y^2} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)
Ax+bydydx=0            ...(2)Ax + by{{dy} \over {dx}} = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)
A+Byd2ydx2+B(dydx)2=0            ...(3)A + By{{{d^2}y} \over {d{x^2}}} + B{\left( {{{dy} \over {dx}}} \right)^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( 3 \right)

From

(2)(2)

and

(3)(3)
x{−Byd2ydx2−B(dydx)2}+Bydydx=0x\left\{ { - By{{{d^2}y} \over {d{x^2}}} - B{{\left( {{{dy} \over {dx}}} \right)}^2}} \right\} + By{{dy} \over {dx}} = 0

Dividing both sides by

−B,-B,

we get

⇒xyd2ydx2+x(dydx)2−ydydx=0\Rightarrow xy{{{d^2}y} \over {d{x^2}}} + x{\left( {{{dy} \over {dx}}} \right)^2} - y{{dy} \over {dx}} = 0

Which is

DEDE

of order

22

and degree

1.1.
Q12
The differential equation of all circles passing through the origin and having their centres on the xx-axis is :
A y2=x2+2xydydx{y^2} = {x^2} + 2xy{{dy} \over {dx}}
B y2=x2−2xydydx{y^2} = {x^2} - 2xy{{dy} \over {dx}}
C x2=y2+xydydx{x^2} = {y^2} + xy{{dy} \over {dx}}
D x2=y2+3xydydx{x^2} = {y^2} + 3xy{{dy} \over {dx}}
Correct Answer
Option A
Solution

General equation of circles passing through origin and having their centres on the

xx

-axis is

x2+y2+2gx=0    ....(i){x^2} + {y^2} + 2gx = 0\,\,\,\,....\left( i \right)

On differentiating

w.r.t.x,w.r.t.x,

we get

2x+2y.dydx+2g=02x + 2y.{{dy} \over {dx}} + 2g = 0
⇒g=−(x+ydydx)\Rightarrow g = - \left( {x + y{{dy} \over {dx}}} \right)

∴\therefore equation

(i)(i)

be

x2+y2+2{−(x+ydydx)}x=0{x^2} + {y^2} + 2\left\{ { - \left( {x + y{{dy} \over {dx}}} \right)} \right\}x = 0
⇒x2+y2−2x2−2xdydx.y=0\Rightarrow {x^2} + {y^2} - 2{x^2} - 2x{{dy} \over {dx}}.y = 0
⇒y2=x2+2xydydx\Rightarrow {y^2} = {x^2} + 2xy{{dy} \over {dx}}
Q13
The solution of the differential equation dydx=x+yx{{dy} \over {dx}} = {{x + y} \over x} satisfying the condition y(1)=1y(1)=1 is :
A y=ln⁡x+xy = \ln x + x
B y=xln⁡x+x2y = x\ln x + {x^2}
C y=xe(x−1) y = x{e^{\left( {x - 1} \right)}}\,
D y=x ln⁡x+xy = x\,\ln x + x
Correct Answer
Option D
Solution
dydx=x+yx=1+yx{{dy} \over {dx}} = {{x + y} \over x} = 1 + {y \over x}

Putting

y=y=
vxvx

and

dydx=v+xdvdx{{dy} \over {dx}} = v + x{{dv} \over {dx}}

we get

v+xdvdx=1+vv + x{{dv} \over {dx}} = 1 + v
⇒∫dxx=∫dv\Rightarrow \int {{{dx} \over x}} = \int {dv}
⇒v=ln⁡ x+c\Rightarrow v = \ln {\mkern 1mu} x + c
⇒y=xln⁡x+cx\Rightarrow y = x\ln x + cx

As

    y(1)=1\,\,\,\,y\left( 1 \right) = 1

∴\therefore

c=1c=1

So solution is

y=xlnx+xy=xlnx+x
Q14
The differential equation which represents the family of curves y=c1ec2x,y = {c_1}{e^{{c_2}x}}, where c1{c_1} , and c2{c_2} are arbitrary constants, is
A y′′=y′yy'' = y'y
B yy′′=y′yy'' = y'
C yy′′=(y′)2yy'' = {\left( {y'} \right)^2}
D y′=y2y' = {y^2}
Correct Answer
Option C
Solution

We have

y=c1ec2xy = {c_1}{e^{{c_2}x}}
⇒y′=c1c2ec2x=c2y\Rightarrow y' = {c_1}{c_2}{e^{{c_2}x}} = {c_2}y
⇒y′y=c2\Rightarrow {{y'} \over y} = {c_2}
⇒y′′y(y′)2y2=0\Rightarrow {{y''y\left( {y'} \right){}^2} \over {{y^2}}} = 0
⇒y′′y=(y′)2\Rightarrow y''y = {\left( {y'} \right)^2}
Q15
The differential equation of the family of curves, x2 = 4b(y + b), b ∈ \in R, is :
A x(y')2 = x – 2yy'
B x(y')2 = 2yy' – x
C xy" = y'
D x(y')2 = x + 2yy'
Correct Answer
Option D
Solution

x2 = 4b(y + b) ⇒\Rightarrow 2x = 4by' ⇒\Rightarrow b =

x2y′{x \over {2y'}}

∴\therefore differential equation is x2 = 4.y.

x2y′{x \over {2y'}}

+ 4

(x2y′)2{\left( {{x \over {2y'}}} \right)^2}

⇒\Rightarrow x2 =

2xyy′{{2xy} \over {y'}}

+

x2(y′)2{{{x^2}} \over {{{\left( {y'} \right)}^2}}}

⇒\Rightarrow x =

2yy′{{2y} \over {y'}}

+

x(y′)2{x \over {{{\left( {y'} \right)}^2}}}

⇒\Rightarrow x(y')2 = x + 2yy'

Q16
Solution of the differential equation cos⁡x dy=y(sin⁡x−y)dx,  0<x<π2\cos x\,dy = y\left( {\sin x - y} \right)dx,\,\,0 < x <{\pi \over 2} is :
A ysec⁡x=tan⁡x+cy\sec x = \tan x + c
B ytan⁡x=sec⁡x+cy\tan x = \sec x + c
C tan⁡x=(sec⁡x+c)y\tan x = \left( {\sec x + c} \right)y
D sec⁡x=(tan⁡x+c)y\sec x = \left( {\tan x + c} \right)y
Correct Answer
Option D
Solution
cos⁡xdy=y(sin⁡x−y)dx\cos xdy = y\left( {\sin x - y} \right)dx
dydx=ytan⁡x−y2  sec⁡ x{{dy} \over {dx}} = y\tan x - {y^2}\,\,\sec \,x
1y2dydx−1y tan⁡ x=−sec⁡ x ....(i){1 \over {{y^2}}}{{dy} \over {dx}} - {1 \over y}\,\tan \,x = - \sec \,x\,....\left( i \right)

Let

    1y=t⇒−1y2dydx=dtdx\,\,\,\,{1 \over y} = t \Rightarrow - {1 \over {{y^2}}}{{dy} \over {dx}} = {{dt} \over {dx}}

From equation

(i)(i)
−dtdx−tan⁡x=−sec⁡x- {{dt} \over {dx}} - \tan x = - \sec x
⇒dtdx+(tan⁡x)t=sec⁡x\Rightarrow {{dt} \over {dx}} + \left( {\tan x} \right)t = \sec x
I.F.=e∫tan⁡xdx=(e)log⁡∣sec⁡x∣sec⁡x{\rm I}.F. = {e^{\int {\tan xdx} }} = {\left( e \right)^{\log \left| {\sec x} \right|}}\sec x

Solution

::
  t(I.F)=∫(I.F)sec⁡xdx\,\,t\left( {{\rm I}.F} \right) = \int {\left( {{\rm I}.F} \right)\sec xdx}
⇒1ysec⁡x=tan⁡x+c\Rightarrow {1 \over y}\sec x = \tan x + c
Q17
Let II be the purchase value of an equipment and V(t)V(t) be the value after it has been used for tt years. The value V(t)V(t) depreciates at a rate given by differential equation dv(t)dt=−k(T−t),{{dv\left( t \right)} \over {dt}} = - k\left( {T - t} \right), where k>0k>0 is a constant and TT is the total life in years of the equipment. Then the scrap value V(T)V(T) of the equipment is
A I−kT22I - {{k{T^2}} \over 2}
B I−k(T−t)22I - {{k{{\left( {T - t} \right)}^2}} \over 2}
C e−kT{e^{ - kT}}
D T2−1k{T^2} - {1 \over k}
Correct Answer
Option A
Solution
dV(t)dt=−k(T−t){{dV\left( t \right)} \over {dt}} = - k\left( {T - t} \right)
⇒∫dVt=−k∫(T−t)dt\Rightarrow \int {dVt} = - k\int {\left( {T - t} \right)} dt
V(t)=k(T−t)22+cV\left( t \right) = {{k{{\left( {T - t} \right)}^2}} \over 2} + c
V(0)=I⇒I=KT22+CV\left( 0 \right) = I \Rightarrow I = {{K{T^2}} \over 2} + C
⇒C=I−KT22\Rightarrow C = I - {{K{T^2}} \over 2}

∴\therefore

     V(T)=0+C=I−KT22\,\,\,\,\,V\left( T \right) = 0 + C = I - {{K{T^2}} \over 2}
Q18
If dydx=y+3>0  {{dy} \over {dx}} = y + 3 > 0\,\, and y(0)=2,y(0)=2, then y(ln⁡2)y\left( {\ln 2} \right) is equal to :
A 55
B 1313
C −2-2
D 77
Correct Answer
Option D
Solution
dydx=y+3⇒∫dyy+3=∫dx{{dy} \over {dx}} = y + 3 \Rightarrow \int {{{dy} \over {y + 3}}} = \int {dx}
⇒ℓn∣y+3∣=x+c\Rightarrow \ell n\left| {y + 3} \right| = x + c

Since

y(0)=2,   y\left( 0 \right) = 2,\,\,\,
       \,\,\,\,\,\,\,

∴\therefore

       \,\,\,\,\,\,\,
ℓn5=c\ell n5 = c
⇒ℓn∣y+3=x+ℓn5∣\Rightarrow \ell n\left| {y + 3 = x + \ell n5} \right|

When

x=ℓn2,x = \ell n2,

then

ℓn∣y+3∣=ℓn2+ℓn5\ell n\left| {y + 3} \right| = \ell n2 + \ell n5
⇒ln⁡∣y+3∣=ℓn 10\Rightarrow \ln \left| {y + 3} \right| = \ell n\,10

∴\therefore

     \,\,\,\,\,
y+3=±10⇒y=7,−13y + 3 = \pm 10 \Rightarrow y = 7, - 13
Q19
The population pp (t)(t) at time tt of a certain mouse species satisfies the differential equation dp(t)dt=0.5 p(t)−450.  {{dp\left( t \right)} \over {dt}} = 0.5\,p\left( t \right) - 450.\,\, If p(0)=850,p(0)=850, then the time at which the population becomes zero is :
A 2ln2ln 1818
B lnln 99
C 12{1 \over 2}lnln 1818
D lnln 1818
Correct Answer
Option A
Solution

Given differential equation is

dp(t)dt=0.5p(t)−450{{dp\left( t \right)} \over {dt}} = 0.5p\left( t \right) - 450
⇒dp(t)dt=12p(t)−450\Rightarrow {{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 450
⇒dp(t)dt=p(t)−9002\Rightarrow {{dp\left( t \right)} \over {dt}} = {{p\left( t \right) - 900} \over 2}
⇒2dp(t)dt=[900−p(t)]\Rightarrow 2{{dp\left( t \right)} \over {dt}} = \left[ {900 - p\left( t \right)} \right]
⇒2dp(t)900−p(t)=−dt\Rightarrow 2{{dp\left( t \right)} \over {900 - p\left( t \right)}} = - dt

Integrate both sides, we get

                −2∫dp(t)900−p(t)=∫dt\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2\int {{{dp\left( t \right)} \over {900 - p\left( t \right)}}} = \int {dt}
                Let  900−p(t)=u\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Let\,\,900 - p\left( t \right) = u
⇒−dp(t)=du\Rightarrow - dp\left( t \right) = du

∴\therefore

      \,\,\,\,\,\,

We have,

2∫duu=∫dt⇒2 ln⁡ u=t+c2\int {{{du} \over u}} = \int {dt \Rightarrow 2\,\ln \,u = t + c}
⇒2ln⁡[900−p(t)]=t+c\Rightarrow 2\ln \left[ {900 - p\left( t \right)} \right] = t + c
              \,\,\,\,\,\,\,\,\,\,\,\,\,\,

when

t=0,p(0)=850t = 0,p\left( 0 \right) = 850
              \,\,\,\,\,\,\,\,\,\,\,\,\,\,
2ln⁡(50)=c2\ln \left( {50} \right) = c
⇒2[ln⁡(900−p(t)50)]=t\Rightarrow 2\left[ {\ln \left( {{{900 - p\left( t \right)} \over {50}}} \right)} \right] = t
⇒900−p(t)=50et2\Rightarrow 900 - p\left( t \right) = 50{e^{{t \over 2}}}
⇒p(t)=900−50et2\Rightarrow p\left( t \right) = 900 - 50{e^{{t \over 2}}}

Let

p(t1)=0p\left( {{t_1}} \right) = 0
0=900−50et120 = 900 - 50{e^{{{{t_1}} \over 2}}}

∴\therefore

      t1=2ln⁡18\,\,\,\,\,\,{t_1} = 2\ln 18
Q20
At present, a firm is manufacturing 20002000 items. It is estimated that the rate of change of production P w.r.t. additional number of workers xx is given by dpdx=100−12x.{{dp} \over {dx}} = 100 - 12\sqrt x . If the firm employs 2525 more workers, then the new level of production of items is
A 25002500
B 30003000
C 35003500
D 45004500
Correct Answer
Option C
Solution

Given, Rate of change is

dpdx=100−12x{{dp} \over {dx}} = 100 - 12\sqrt x
⇒dP=(100−12x)dx\Rightarrow dP = \left( {100 - 12\sqrt x } \right)dx

By intergrating

∫dP=∫(100−12x)dx\int {dP = \int {\left( {100 - 12\sqrt x } \right)} } dx
∫dP=∫(100−12x)dx\int {dP} = \int {\left( {100 - 12\sqrt x } \right)} dx
P=100x−8x3/2+CP = 100x - 8{x^{3/2}} + C

Given, when

x=0x=0

then

P=2000P=2000
⇒C=2000\Rightarrow C = 2000

Now when

xx
=25=25

then

P=100×25−8×(25)3/2+2000P = 100 \times 25 - 8 \times {\left( {25} \right)^{3/2}} + 2000
          \,\,\,\,\,\,\,\,\,\,
=4500−1000=4500-1000
⇒P=3500\Rightarrow P = 3500
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