12th NCERT Differential Equation Exercise 9.4 Number of questions 23
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For each of the differential equations in Exercises 1 to 10, find the general solution

Question (1)

$\frac{{dy}}{{dx}} = \frac{{1 - \cos x}}{{1 + \cos x}}$

Solution

$dy = \frac{{2{{\sin }^2}\frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}dx$ dy = tan2 (x/2)dx
∫ dy = ∫ tan2 (x/2)dx
$\int {dy = \int {\left( {{{\sec }^2}\frac{x}{2} - 1} \right)} dx}$ $\int {dy = \int {{{\sec }^2}\frac{x}{2}dx - \int {dx} } }$ $y = \frac{{\tan \frac{x}{2}}}{{\frac{1}{2}}} - x + c$ $y = 2\tan \frac{x}{2} - x + C$

Question (2)

$\frac{{dy}}{{dx}} = \sqrt {4 - {y^2}} \left( { - 2 < y < 2} \right)$

Solution

$\frac{{dy}}{{\sqrt {4 - {y^2}} }} = dx$ $\int {\frac{{dy}}{{\sqrt {4 - {y^2}} }}} = \int {dx}$ ${\sin ^{ - 1}}\frac{y}{2} = x + C$ y= 2sin(x+C)

Question (3)

$\frac{{dy}}{{dx}} + y = 1\left( {y \ne 1} \right)$

Solution

$\frac{{dy}}{{dx}} = 1 - y$ Seperation of variable
$\frac{{dy}}{{1 - y}} = dx$ $\int {\frac{{dy}}{{1 - y}}} = \int {dx}$ $\frac{{\log \left| {1 - y} \right|}}{{ - 1}} = x + c$ log|1-y| = -x + c
1 - y= e-x+c
1 - y=e-x . ec
1 - y = ece-x
1 - ece-x = y
Let - ec = A
y = 1 + A e-x

Question (4)

sec2 x tany dx + sec2y tanx dy = 0

Solution

∴ sec2y tanx dy = -sec2 tany dx $\frac{{{{\sec }^2}ydy}}{{\tan y}} = \frac{{ - {{\sec }^2}xdx}}{{\tan x}}$ Let tany = u
sec2y dy = du
and tanx = v
sec2xdx = dv $\int {\frac{{du}}{u} = - \int {\frac{{dv}}{v}} }$ log u = -log v + log c
log u + log v = log c
log(uv) = logc
uv = c
∴ tany tanx = c

Question (5)

( ex + e-x)dy - (ex - e-x) dx = 0

Solution

( ex + e-x)dy = (ex - e-x) dx $dy = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}dx$ $\int {dy} = \int {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}dx}$ Let ex + e-x = t
∴ [ex + e-x(-1)] = dt
(ex - e-x) dx = dt $\int {dy} = \int {\frac{{dt}}{t}}$ y = log|t| + c
y = log|ex + e-x|+c

Question (6)

$\frac{{dy}}{{dx}} = \left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)$

Solution

$\frac{{dy}}{{\left( {1 + {y^2}} \right)}} = \left( {1 + {x^2}} \right)dx$ $\int {\frac{{dy}}{{\left( {1 + {y^2}} \right)}}} = \left( {1 + {x^2}} \right)dx$ tan-1y = x + x3/3 + c

Question (7)

y log y dx - x dy = 0

Solution

$\frac{{dy}}{{y\log y}} = \frac{{dx}}{x}$ $\int {\frac{{dy}}{{y\log y}}} = \int {\frac{{dx}}{x}}$ Let log y = t
dy/y = dt $\int {\frac{{dt}}{t}} = \log x + \log c$ log t = log (cx)
∴ t = cx
log y = cx
y = ecx

Question (8)

${x^5}\frac{{dy}}{{dx}} = - {y^5}$

Solution

x5dy = -y5 dx $\frac{{dy}}{{{y^5}}} = - \frac{{dx}}{{{x^5}}}$ $\int {{y^{ - 5}}dy = - \int {{x^{ - 5}}dx} }$ $\frac{{{y^{ - 4}}}}{{ - 4}} = - \left( {\frac{{ {x^4}}}{{ - 4}}} \right) + c$ $\frac{1}{{{y^4}}} = \frac{{ - 1}}{{{x^4}}} + c$ $\frac{1}{{{x^4}}} + \frac{1}{{{y^4}}} = c$

Question (9)

$\frac{{dy}}{{dx}} = {\sin ^{ - 1}}x$

Solution

dy = sin-1x dx
Here sin-1x = u and v=1
Using integration by parts $\int {uvdx = u\int {vdx} - \int {\frac{d}{{dx}}u\int {vdx} } }$
$y = {\sin ^{ - 1}}x\int {dx} - \int {\left( {\frac{d}{{dx}}{{\sin }^{ - 1}}x\int {dx} } \right)} dx$ $y = x{\sin ^{ - 1}}x - \int {\frac{1}{{\sqrt {1 - {x^2}} }}xdx}$ Let 1 - x2 = t2
-2x dx = 2t dt
xdx = -t dt $y = x{\sin ^{ - 1}}x - \int {\frac{{ -\require{cancel}\cancel{t}}}{\cancel{t}}dt}$ y= xsin-1x + t + c $y = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + c$

Question (10)

ex tan y dx + ( 1- ex)sec2y dy = 0

Solution

(1-ex) sec2 y dy = - ex tan y dx $\frac{{{{\sec }^2}y}}{{\tan y}}dy = \frac{{ - {e^x}}}{{1 - {e^x}}}dx$ $\int {\frac{{{{\sec }^2}y}}{{\tan y}}dy} = \int {\frac{{ - {e^x}}}{{1 - {e^x}}}dx}$ Let tan y = t sec2y dy = dt
1-ex = u
-ex dx = du
$\int {\frac{{dt}}{t} = \int {\frac{{du}}{u}} }$ log t = log u + log c
∴ log t = log (uc)
∴ t = cu
∴ tan y = c (1 - ex)

For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition:

Question (11)

$\left( {{x^3} + {x^2} + x + 1} \right)\frac{{dy}}{{dx}} = 2{x^2} + x$ y =1 when x = 0

Solution

$dy = \frac{{2{x^2} + x}}{{\left( {{x^3} + {x^2} + x + 1} \right)}}dx$ $dy = \frac{{2{x^2} + x}}{{{x^2}\left( {x + 1} \right) + \left( {x + 1} \right)}}dx$ $dy = \frac{{2{x^2} + x}}{{\left( {{x^2} + 1} \right)\left( {x + 1} \right)}}dx$ $\int {dy} = \int {\frac{{2{x^2} + x}}{{\left( {{x^2} + 1} \right)\left( {x + 1} \right)}}dx}$ y =I
$\text{where}\qquad I = \int {\frac{{2{x^2} + x}}{{\left( {{x^2} + 1} \right)\left( {x + 1} \right)}}dx}$ $\text{let}\qquad \frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} + 1}} = \frac{{2{x^2} + x}}{{\left( {{x^2} + 1} \right)\left( {x + 1} \right)}}$ ∴ A(x2+1) + (Bx+C)(x+1) = 2x2+x
If x = -1 ⇒ 2A = 1 ⇒ A= 1/2
If x=0 ⇒ A+C = 0 ⇒ C = -1/2
If x=1 ⇒ 2A+2B+2C = 3
1+2B-1 = 3 ⇒ B = 3/2
$\text{let}\qquad I = \int {\left( {\frac{{\frac{1}{2}}}{{x + 1}} + \frac{{\frac{3}{2}x - \frac{1}{2}}}{{{x^2} + 1}}} \right)dx}$ $I = \frac{1}{2}\int {\frac{{dx}}{{x + 1}} + \frac{3}{2}\int {\frac{{xdx}}{{{x^2} + 1}} - \frac{1}{2}\int {\frac{{dx}}{{{x^2} + 1}}} } }$ Let x2 + 1 = u
2xdx = du
xdx= du/2
$I = \frac{1}{2}\log \left| {x + 1} \right| + \frac{3}{2}\int {\frac{{\frac{{du}}{2}}}{u} - \frac{1}{2}{{\tan }^{ - 1}}x + c}$ $I = \frac{1}{2}\log \left| {x + 1} \right| + \frac{3}{4}\log u - \frac{1}{2}{\tan ^{ - 1}}x + c$ $I = \frac{1}{2}\log \left| {x + 1} \right| + \frac{3}{4}\log \left| {{x^2} + 1} \right| - \frac{1}{2}{\tan ^{ - 1}}x + c$ $I = \frac{1}{4}\left[ {2\log \left| {x + 1} \right| + 3\log \left| {{x^2} + 1} \right|} \right] - \frac{1}{2}{\tan ^{ - 1}}x + c$ $I = \frac{1}{4}\log \left| {{{\left( {x + 1} \right)}^2}{{\left( {{x^2} + 1} \right)}^3}} \right| - \frac{1}{2}{\tan ^{ - 1}}x + c$ Now y =I
$y = \frac{1}{4}\left[ {\log {{\left( {x + 1} \right)}^2}{{\left( {{x^2} + 1} \right)}^3}} \right] - \frac{1}{2}{\tan ^{ - 1}}x + c$ y =1 when x = 0 $1 = \frac{1}{4}\left[ {\log \left( 1 \right)} \right] - \frac{1}{2}{\tan ^{ - 1}}0 + c$ 1 = 0 - 0 + c
c =1 $y = \frac{1}{4}\left[ {\log {{\left( {x + 1} \right)}^2}{{\left( {{x^2} + 1} \right)}^3}} \right] - \frac{1}{2}{\tan ^{ - 1}}x + 1$

Question (12)

$x\left( {{x^2} - 1} \right)\frac{{dy}}{{dx}} = 1$ y = 0 when x = 2

Solution

$dy = \frac{{dx}}{{x\left( {{x^2} - 1} \right)}}$ $\int {dy} = \int {\frac{{dx}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}}$ y = I $\text{Let}\qquad \frac{A}{x} + \frac{B}{{x - 1}} + \frac{C}{{x + 1}} = \frac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}$ A(x-1)(x+1) + Bx(x+1) + Cx(x-1) = 1
If x= 0 ⇒ -A = 1 ⇒ A = -1
If x=1 ⇒ 2B = 1 ⇒ B = 1/2
If x = -1 ⇒ C(-1)(-2) = 1 ⇒ C=1/2 $y = \int {\left( {\frac{{ - 1}}{x} + \frac{{\frac{1}{2}}}{{x - 1}} + \frac{{\frac{1}{2}}}{{x + 1}}} \right)dx}$ $y = - \int {\frac{1}{x}dx + \frac{1}{2}\int {\frac{1}{{x - 1}}dx} + \frac{1}{2}\int {\frac{1}{{x + 1}}dx} }$ $y = - \log x + \frac{1}{2}\log \left| {x - 1} \right| + \frac{1}{2}\log \left| {x + 1} \right| + c$ $y = \frac{1}{2}\left[ {\log \left( {x - 1} \right)\left( {x + 1} \right) - 2\log x} \right]$ $y = \frac{1}{2}\log \left| {\frac{{{x^2} - 1}}{{{x^2}}}} \right| + c$ when y=0, x=2 $0 = \frac{1}{2}log\left| {\frac{3}{4}} \right| + c$ $c = - \frac{1}{2}\log \left| {\frac{3}{4}} \right|$ $y = \frac{1}{2}\log \left| {\frac{{{x^2} - 1}}{{{x^2}}}} \right| - \frac{1}{2}\log \left| {\frac{3}{4}} \right|$

Question (13)

$\cos \left( {\frac{{dy}}{{dx}}} \right) = a$ (a∈R) : y = 1 when x = 0

Solution

$\frac{{dy}}{{dx}} = {\cos ^{ - 1}}a$ $dy = {\cos ^{ - 1}}adx$ $\int {dy} = {\cos ^{ - 1}}a\int {dx}$ y = cos-1ax + c
when x = 0, y = 1
1 = cos-1a (0) + c
c = 1
y = xcos-1a + 1
$\frac{{y - 1}}{x} = {\cos ^{ - 1}}a$ $\cos \left( {\frac{{y - 1}}{x}} \right) = a$

Question (14)

$\frac{{dy}}{{dx}} = y\tan x$ y = 1, x = 0

Solution

$\frac{{dy}}{y} = \tan xdx$ $\int {\frac{{dy}}{y}} = \int {\tan xdx}$ log y = log|secx| + logc
log y = log c(secx)
y = c(secx)
y=1 when x= 0
1 = c sec0 ⇒ c = 1
y = (secx)

Question (15)

Find the equation of a curve passing through the point (0, 0) and whose differential equation is y' = exsin x

Solution

curve passes through (0, 0) equation is
y' = ex sin x
$\frac{{dy}}{{dx}} = {e^x}\sin x$ $dy = \sin x{e^x}dx$ $\int {dy} = \int {\sin x{e^x}dx}$ Using integration by parts
Let u = sinx and v = ex $I = \sin x\int {{e^x}dx - \int {\left( {\frac{d}{{dx}}\sin x\int {{e^x}dx} } \right)} } dx$ $I = \sin x{e^x} - \int {\cos x{e^x}dx}$ $I = \sin x{e^x} - \left[ {\cos x\int {{e^x}dx - \int {\left( {\frac{d}{{dx}}\cos x\int {{e^x}dx} } \right)dx} } } \right]$ I= sinxex - cosxexdx + ∫ -sinx exdx
I = (sinx - cosx)ex - I
2I = (sinx - cosx) ex + c
$I = \left( {\frac{{\sin x - \cos x}}{2}} \right){e^x} + c$ Now y = I
$y = \frac{{{e^x}}}{2}\left( {\sin x - cosx} \right) + c$ it passes through (0, 0) $0 = \frac{{{e^0}}}{2}\left( {\sin 0 - \cos 0} \right) + c$ 0 = ½(0 - 1)+c
c= ½ $y = \frac{{{e^x}}}{2}\left( {\sin x - \cos x} \right) + \frac{1}{2}$ 2y - 1 = ex ( sinx-cosx)

Question (16)

For the differential equation $xy\frac{{dy}}{{dx}} = \left( {x + 2} \right)\left( {y + 2} \right)$ find the solution curve passing through the point (1, -1)

Solution

$xy\frac{{dy}}{{dx}} = \left( {x + 2} \right)\left( {y + 2} \right)$ $\frac{y}{{y + 2}}dy = \frac{{x + 2}}{x}dx$ $\int {\frac{y}{{y + 2}}dy} = \int {\frac{{x + 2}}{x}dx}$ $\int {\frac{{y + 2 - 2}}{{y + 2}}dy} = \int {dx + 2\int {\frac{1}{x}dx} }$ $\int {dy - 2\int {\frac{{dy}}{{y + 2}}} } = \int {dx + 2\int {\frac{1}{x}dx} }$ $y - 2\log \left| {y + 2} \right| = x + 2\log x + c$ y -2 log |y+2| = x + 2log x + c
y-x = log x2 + log (y+2)2 + c
y-x = log |x2 (y+2)2| + c
it passes through ( 1, -1)
-1-1 = log|1(1)|+ c
c= -2
y-x = log |x2 (y+2)2| - 2
y-x + 2 = log |x2 (y+2)2|

Question (17)

Find the equation of a curve passing through the point (0, -2) given that at any point (x, y) on the curve, the product of the slope of its tanget and y-coordinate of the point is equal to the x-coordinate of the point

Solution

Curve passes through (0, -2)
Slope of the tangent × y-coordinate = x coordinate
$\frac{{dy}}{{dx}}y = x$ $ydy = xdx$ $\int {ydy} = \int {xdx}$ $\frac{{{y^2}}}{2} = \frac{{{x^2}}}{2} + c$ it passes through(0, -2)
4/2 = 0 +c ⇒ c = 2 $\frac{{{y^2}}}{2} = \frac{{{x^2}}}{2} + 2$ y2 = x2 + 4
∴ y2 - x2 = 4

Question (18)

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point ( -4, -3). Find the equation of the curve given that it passes through (-2, 1)

Solution

Slope of tangent = 2 × slope of line segment AB. A(x, y), B(-4, 3)
$\frac{{dy}}{{dx}} = 2 \times \frac{{y + 3}}{{x + 4}}$ $\frac{{dy}}{{y + 3}} = 2 \times \frac{{dx}}{{x + 4}}$ $\int {\frac{{dy}}{{y + 3}}} = 2 \times \int {\frac{{dx}}{{x + 4}}}$ log |y+3| = 2log|x+4| + log c
log|y+3| = log c(x+4)2
⇒ y+3 = c ( x + 4)2
it passes the (-2, 1)
4 = c (4)
c = 1
∴ y +3 = (x+4)2

Question (19)

The volume of spherical ballon being inflated changes at a constant rate. If initally its radius is 3 units and after 3 seconds it is 6 units. Find the radius of ballon after 't' seconds

Solution

Volume of sphere V
$V = \frac{4}{3}\pi {r^3}$ It is given rate of change of volume is constant $\frac{{dV}}{{dt}} = K$ $\frac{d}{{dt}}\left( {\frac{4}{3}\pi {r^3}} \right) = K$ $\frac{4}{\cancel{3}}\pi \cdot \cancel{3}{r^2}\frac{{dr}}{{dt}} = K$ $4\pi {r^2}dr = Kdt$ $4\pi \int {{r^2}dr} = K\int {dt}$ $\frac{{4\pi }}{3}{r^3} = Kt + c$ $4\pi {r^3} = 3Kt + 3c$ When t = 0 , r = 3
4π (3)3 = 3K(0) + 3c
3c = 108π
c = 36π
Where t = 3, r = 6
4π(6)3 3[K(3) + 36π]
864π = 9K +108π
$K = \frac{{864\pi - 108\pi }}{9}$
K = 84π
4πr3 = 3(84π)t + 108π
4r3 = 252 t + 108
r3 = (63t + 27)1/3

Question (20)

In bank, principal increases continuously at the rate of 5% per year. Find the value of r if Rs100 double itself in 10 years (loge2=0.6931)

Solution

Let P is principal, r = rate $\frac{{dP}}{{dt}} = \left( {\frac{r}{{100}}} \right)P$ $\frac{{dP}}{P} = \frac{r}{{100}}dt$ $\int {\frac{{dP}}{P}} = \frac{r}{{100}}\int {dt}$ $\log P = \frac{r}{{100}}t + \log c$ ${\log _e}\left( {\frac{P}{c}} \right) = \frac{{rt}}{{100}}$ $\frac{P}{c} = {e^{\frac{{rt}}{{100}}}}$ $P = c{e^{\frac{{rt}}{{100}}}}$ When t=0, P=100
100 = c e0
c = 100
$P = 100{e^{\frac{{rt}}{{100}}}}$ P = 200 where t =10 $200 = 100{e^{\frac{{10r}}{{100}}}}$ $2 = {e^{\frac{r}{{10}}}}$ ${\log _e}2 = \frac{r}{{10}}$ $0.6931 = \frac{r}{{10}}$ r = 6.931

Question (21)

In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs1000 is depositedd with this bank, how much will it worth after 10 years (e0.5 = 1.648)

Solution

P = Principal amount $\frac{{dP}}{{dt}} = \frac{5}{{100}}P$ $\frac{{dP}}{P} = \frac{5}{{100}}dt$$\int {\frac{{dP}}{P}} = \frac{5}{{100}}\int {dt}$ $\int {\frac{{dP}}{P}} = \frac{5}{{100}}\int {dt}$ $\log P = \frac{5}{{100}}t + \log c$ $\log \frac{P}{c} = \frac{5}{{100}}t$ $\frac{P}{c} = {e^{\frac{5}{{100}}t}}$ $P = c{e^{\frac{5}{{100}}t}}$ When t = 0, P=1000 $1000 = c{e^{\frac{5}{{100}}t}}$ c= 1000 $P = 1000{e^{\frac{5}{{100}}t}}$ t=10, P = ? $P = 1000{e^{\frac{5}{{\cancel{100}^{10}}} \times \cancel{10}}}$ $P = 1000{e^{0.5}}$ P= 1000 × 1.648
P = 1648

Question (22)

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Solution

Let x be the number of bacteria present at that time $\frac{{dx}}{{dt}} \propto x$ $\frac{{dx}}{{dt}} = kx$ $\frac{{dx}}{x} = kdt$ $\int {\frac{{dx}}{x}} = k\int {dt}$ logx = kt + logc $\log \left( {\frac{x}{c}} \right) = kt$ $\frac{x}{c} = {e^{kt}}$ $x = c{e^{kt}}$ When t = 0, x=1,00,000
1,00,000 = cek(0)
1,00,000 = c
x= 1,00,000 ekt
In t = 2hr , 10% of 1,00,000 = 10,000
x = 10,000 + 1,00,000 = 1,10,000
1,10,000 = 1,00,000 e2k
1.1 = e2k
2k = loge(1.1
x=2,00,000, t =?
2,00,000 = 1,00,000 ekt
2 = ekt
kt = log2
$\frac{{{{\log }_e}1.1}}{2}t = \log 2$ $t = \frac{{2\log 2}}{{\log 1.1}}$

Question (23)

The general solution of the differential equation $\frac{{dy}}{{dx}} = {e^{x + y}}$ (A) ex + e-y = C
(B) ex + ey = C
(C) e-x + ey = C
(D) e-x + e-y = C

Solution

$\frac{{dy}}{{dx}} = {e^x} \cdot {e^y}$ $\frac{{dy}}{{{e^y}}} = {e^x}dx$ $\Rightarrow \int {{e^{ - y}}dy} = \int {{e^x}dx}$ $\frac{{{e^{ - y}}}}{{ - 1}} = {e^x} + {c_1}$ -e-y = ex + c1
ex + e-y = -c1
ex + e-y = C
Option (A) is correct