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 = tan
2 (x/2)dx
∫ dy = ∫ tan
2 (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 . e
c
1 - y = e
ce
-x
1 - e
ce
-x = y
Let - e
c = A
y = 1 + A e
-x
Question (4)
sec
2 x tany dx + sec
2y tanx dy = 0
Solution
∴ sec
2y tanx dy = -sec
2 tany dx
\[\frac{{{{\sec }^2}ydy}}{{\tan y}} = \frac{{ - {{\sec }^2}xdx}}{{\tan x}}\]
Let tany = u
sec
2y dy = du
and tanx = v
sec
2xdx = 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)
( e
x + e
-x)dy - (e
x - e
-x) dx = 0
Solution
( e
x + e
-x)dy = (e
x - 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 e
x + e
-x = t
∴ [e
x + e
-x(-1)] = dt
(e
x - e
-x) dx = dt
\[\int {dy} = \int {\frac{{dt}}{t}} \]
y = log|t| + c
y = log|e
x + 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 + x
3/3 + 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 = e
cx
Question (8)
\[{x^5}\frac{{dy}}{{dx}} = - {y^5}\]
Solution
x
5dy = -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 - x
2 = t
2
-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)
e
x tan y dx + ( 1- e
x)sec
2y dy = 0
Solution
(1-e
x) sec
2 y dy = - e
x 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
sec
2y dy = dt
1-e
x = u
-e
x 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 - e
x)
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(x
2+1) + (Bx+C)(x+1) = 2x
2+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 x
2 + 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' = e
xsin x
Solution
curve passes through (0, 0) equation is
y' = e
x 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 = e
x
\[I = \sin x\int {{e^x}dx - \int {\left( {\frac{d}{{dx}}\sin x\int {{e^x}dx} } \right)} } dx\]
\[I = \sin x{e^x} - \int {\cos x{e^x}dx} \]
\[I = \sin x{e^x} - \left[ {\cos x\int {{e^x}dx - \int {\left( {\frac{d}{{dx}}\cos x\int {{e^x}dx} } \right)dx} } } \right]\]
I= sinxe
x - cosxe
xdx + ∫ -sinx e
xdx
I = (sinx - cosx)e
x - I
2I = (sinx - cosx) e
x + 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 = e
x ( 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 x
2 + log (y+2)
2 + c
y-x = log |x
2 (y+2)
2| + c
it passes through ( 1, -1)
-1-1 = log|1(1)|+ c
c= -2
y-x = log |x
2 (y+2)
2| - 2
y-x + 2 = log |x
2 (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
\[\frac{{{y^2}}}{2} = \frac{{{x^2}}}{2} + 2\]
y
2 = x
2 + 4
∴ y
2 - x
2 = 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
4πr
3 = 3(84π)t + 108π
4r
3 = 252 t + 108
r
3 = (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 (log
e2=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 e
0
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 (e
0.5 = 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 = ce
k(0)
1,00,000 = c
x= 1,00,000 e
kt
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 e
2k
1.1 = e
2k
2k = log
e(1.1
x=2,00,000, t =?
2,00,000 = 1,00,000 e
kt
2 = e
kt
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) e
x + e
-y = C
(B) e
x + e
y = C
(C) e
-x + e
y = 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 = e
x + c
1
e
x + e
-y = -c
1
e
x + e
-y = C
Option (A) is correct