Differential equations are mathematical tools that describe relationships involving rates of change. They are central in many fields such as physics, engineering, economics, and biology, because they effectively model dynamic systems. A differential equation involves a function and its derivatives, expressing how the function changes over time or space.
Question 14. 3x2dy = (3xy + y2)dx
Solution:
We have,
3x2dy = (3xy + y2)dx
(dy/dx) = (3xy + y2)/3x2
It is a homogeneous equation,
So put y = vx (i)
On differentiating both sides w.r.t x,
dy/dx = v + x(dv/dx)
So,
v + x(dv/dx) = (3xvx + v2x2)/3x2
v + x(dv/dx) = (3v + v2)/3
x(dv/dx) = [(3v + v2)/3] - v
x(dv/dx) = (3v + v2 - 3v)/3
3(dv/v2) = (dx/x)
On integrating both sides,
3∫(dv/v2) = ∫(dx/x)
-(3/v) = log|x| + c
-3x/y = log(x) + c (Where ‘c’ is integration constant)
Question 15. (dy/dx) = x/(2y + x)
Solution:
We have,
(dy/dx) = x/(2y + x)
It is a homogeneous equation,
So, put y = vx (i)
On differentiating both sides w.r.t x,
dy/dx = v + x(dv/dx)
So,
v + x(dv/dx) = x/(2vx + x)
v + x(dv/dx) = 1/(2v + 1)
x(dv/dx) = [1/(2v + 1)] - v
x(dv/dx) = (1 - 2v2 - v)/(2v + 1)
(2v + 1)dv/(2v2 + v - 1) = -(dx/x)
On integrating both sides,
∫(2v + 1)dv/(2v2 + v - 1) = -∫(dx/x)
∫\frac{2v+1}{2v(v+1)-1(v+1)}dx=-∫\frac{dx}{x}
∫\frac{2v+1}{(v+1)(2v-1)}dx=-∫\frac{dx}{x} Solving by partial fraction,
\frac{2v+1}{(v+1)(2v-1)}=\frac{A}{(2v-1)}+\frac{B}{(v+1)} A(v + 1) + B(2v - 1) = (2v + 1) (i)
Putting v = -1 and solve above equation,
A(0) + B(-3) = (-1)
B = (1/3)
Putting v = -(1/2) and solve equation (i),
A(3/2) + B(0) = 2
A = (4/3)
\frac{4}{3}∫\frac{dv}{(2v-1)}+\frac{1}{3}∫\frac{dv}{(v+1)}=-log|x|+log|c| (3/2)log|2v - 1| + (1/3)log|v + 1| = -log|x| + log|c|
log|(2v - 1)2(v + 1)| = -log|x|3 + log|c|
|(2v - 1)2(v + 1)| = (c/x3)
(2y/x - 1)2(y/x + 1) = (c/x3)
\frac{(2y-x)^2(x+y)}{x^2x}=\frac{c}{x^3} (2y - x)2(x + y) = c (Where ‘c’ is integration constant)
Question 16. (x + 2y)dx - (2x - y)dy = 0
Solution:
We have,
(x + 2y)dx - (2x - y)dy = 0
(dy/dx) = (x + 2y)/(2x - y)
It is a homogeneous equation,
So, put y = vx (i)
On differentiating both sides w.r.t x,
dy/dx = v + x(dv/dx)
So,
v + x(dv/dx) = (x + 2vx)/(2x - vx)
v + x(dv/dx) = (1 + 2v)/(2 - v)
x(dv/dx) = [(1 + 2v)/(2 - v)] - v
x(dv/dx) = (1 + 2v - 2v + v2)/(2 - v)
(2 - v)dv/(1 + v2) = (dx/x)
On integrating both sides,
∫(2 - v)dv/(1 + v2) = ∫(dx/x)
2∫dv/(1 + v2) - ∫vdv/(1 + v2) = log|x| + log|c|
2tan-1v - (1/2)∫2vdv/(1 + v2) = log|x| + log|c|
2tan-1v - log|1 + v2|1/2 = log|cx|
2tan-1v = log|cx√(1 + v2)|
e^{2tan^{-1}(\frac{y}{x})}=\frac{(y^2+x^2)^{\frac{1}{2}}}{x}xc
e^{2tan^{-1}(\frac{y}{x})}={(y^2+x^2)^{\frac{1}{2}}}c (Where ‘c’ is integration constant)
Question 17. \frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^2}{x^2}-1}
Solution:
We have,
\frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^2}{x^2}-1} It is a homogeneous equation,
So, put y = vx (i)
On differentiating both sides w.r.t x,
dy/dx = v + x(dv/dx)
So,
v + x(dv/dx) = (vx/x) - √(v2x2/x2 - 1)
v + x(dv/dx) = v - √(v2 - 1)
x(dv/dx) = -√(v2 - 1)
dv/√(v2 - 1) = -(dx/x)
On integrating both sides,
∫dv/√(v2 - 1) = -∫(dx/x)
log|v + √(v2 - 1)| = -log|x| + log|c|
|v + √(v2 - 1)| = (c/x)
\frac{y}{x}-[\sqrt{(\frac{y}{x})^2-1}]x=c y + √(y2 - x2) = c (Where ‘c’ is integration constant)
Question 18. (dy/dx) = (y/x){log(y) - log(x) + 1}
Solution:
We have,
(dy/dx) = (y/x){log(y) - log(x) + 1}
(dy/dx) = (y/x){log(y/x) + 1}
It is a homogeneous equation,
So, put y = vx (i)
On differentiating both sides w.r.t x,
dy/dx = v + x(dv/dx)
So,
v + x(dv/dx) = v{log(v) + 1}
v + x(dv/dx) = vlog(v) + v
x(dv/dx) = vlog(v)
dv/vlogv = (dx/x)
On integrating both sides,
∫dv/vlogv = ∫(dx/x)
Let, logv = z
On differentiating both sides,
dv/v = dz
∫(dz/z) = ∫(dx/x)
log|z| = log|x| + log|c|
z = xc
log|v| = xc
log|y/x| = xc (Where ‘c’ is integration constant)
Question 19. (dy/dx) = (y/x) + sin(y/x)
Solution:
We have,
(dy/dx) = (y/x) + sin(y/x)
It is a homogeneous equation,
So, put y = vx (i)
On differentiating both sides w.r.t x,
dy/dx = v + x(dv/dx)
So,
v + x(dv/dx) = v + sin(v)
x(dv/dx) = sin(v)
dv/sin(v) = (dx/x)
On integrating both sides,
∫dv/sin(v) = ∫(dx/x)
∫cosec(v)dv = ∫(dx/x)
log|tan(v/2)| = log(x) + log(c)
log|tan(y/2x)| = log|xc|
tan(y/2x) = |xc| (Where ‘c’ is integration constant)
Question 20. y2dx + (x2 - xy + y2)dy = 0
Solution:
We have,
y2dx + (x2 - xy + y2)dy = 0
(dy/dx) = -(y2)/(x2 - xy + y2)
It is a homogeneous equation,
So, put y = vx (i)
On differentiating both sides w.r.t x,
dy/dx = v + x(dv/dx)
So,
v + x(dv/dx) = -(v2x2)/(x2 - xvx + v2x2)
v + x(dv/dx) = -(v2)/(1 - v + v2)
y(dv/dx) = [-(v2)/(1 - v + v2)] - v
x\frac{dv}{dx}=\frac{-v^2-v+v^2-v^3}{1-v+v^2}
\frac{1-v+v^2}{-(v+v^3)}dv=\frac{dx}{x}
\frac{1-v+v^2}{-v(1+v^2)}dv=\frac{dx}{x}
(\frac{1}{1+v^2}-\frac{1}{v})dv=\frac{dx}{x} On integrating both sides,
∫dv/(1 + v2) - ∫dv/v = ∫(dx/x)
tan-1(v) - log(v) = log(x) + log(c)
tan-1(y/x) - log|y/x| = log(xc)
tan-1(y/x) = log|(y/x)xc|
tan-1(y/x) = log|yc|
e^{tan^{-1}(\frac{y}{x})}=cy (Where ‘c’ is integration constant)
Question 21. [x√(x2 + y2) - y2]dx + xydy = 0
Solution:
We have,
[x√(x2 + y2) - y2]dx + xydy = 0
dy/dx = -[x√(x2 + y2) - y2]/xy
dy/dx = [y2 - x√(x2 + y2)]/xy
It is a homogeneous equation,
So, put y = vx (i)
On differentiating both sides w.r.t x,
dy/dx = v + x(dv/dx)
So,
v + x(dv/dx) = [v2x2 - x√(x2 + v2x2)]/xvx
v + x(dv/dx) = [v2 - √(1 + v2)]/v
x(dv/dx) = [v2 - √(1 + v2)]/v - v
x\frac{dv}{dx}=\frac{v^2-\sqrt{1+v^2}-v^2}{v} x(dv/dx) = -(√(1 + v2)/v
vdv/√(1 + v2) = -(dx/x)
On integrating both sides,
∫vdv/√(1 + v2) = -∫(dx/x)
(1/2)∫2vdv/√(1 + v2) = -∫(dx/x)
Let, 1 + v2 = z
On differentiating both sides,
2vdv = dz
(1/2)∫dz/√z = -∫(dx/x)
√z = -log|x| + log|c|
√(1 + v2) = log|c/x|
√(x2 + y2)/x = log|c/x|
√(x2 + y2) = xlog|c/x| (Where ‘c’ is integration constant)
Question 22. x(dy/dx) = y - xcos2(y/x)
Solution:
We have,
x(dy/dx) = y - xcos2(y/x)
(dy/dx) = y/x - cos2(y/x)
It is a homogeneous equation,
So, put y = vx (i)
On differentiating both sides w.r.t x,
dy/dx = v + x(dv/dx)
So,
v + x(dv/dx) = v - cos2(v)
x(dv/dx) = -cos2(v)
dv/cos2(v) = -(dx/x)
On integrating both sides,
∫dv/cos2(v) = -∫(dx/x)
∫sec2vdv = -∫(dx/x)
tan(v) = -log|x| + log|c|
tan(y/x) = log|c/x| (Where ‘c’ is integration constant)
Question 23. (y/x)cos(y/x)dx - {(x/y)sin(y/x) + cos(y/x)}dy = 0
Solution:
We have,
(y/x)cos(y/x)dx - {(x/y)sin(y/x) + cos(y/x)}dy = 0
\frac{\frac{y}{x}cos\frac{y}{x}}{\frac{x}{y}sin\frac{y}{x}+cos\frac{y}{x}} It is a homogeneous equation,
So, put y = vx (i)
On differentiating both sides w.r.t x,
dy/dx = v + x(dv/dx)
So,
v+x\frac{dv}{dx}=\frac{vcos(v)}{\frac{1}{v}sin(v)+cos(v)}
x\frac{dv}{dx}=\frac{v^2cos(v)}{sin(v)+vcos(v)}-v x(dv/dx) = (v2cosv - vsinv - v2cosv)/(sinv + vcosv)
x(dv/dx) = -vsinv/(sinv + vcosv)
On integrating both sides,
∫[(sinv + vcosv)/vsinv]dv = -∫(dx/x)
∫(dv/v) + ∫(cotv)dv = -∫(dx/x)
log|v| + log|sinv| = -log|x| + log|c|
log|vsinv| = log|c/x|
(y/x)sin(y/x) = (c/x)
ysin(y/x) = c (Where ‘c’ is integration constant)
Question 24. xylog(x/y)dx + {y2 - x2log(x/y)}dy = 0
Solution:
We have,
xylog(x/y)dx + {y2 - x2log(x/y)}dy = 0
\frac{dx}{dy}=\frac{x^2log(\frac{x}{y})-y^2}{xylog(\frac{x}{y})} It is a homogeneous equation,
So, put x = vy (i)
On differentiating both sides w.r.t y,
dx/dy = v + y(dv/dy)
So,
v+y\frac{dv}{dy}=\frac{v^2y^2log(v)-y^2}{yvylog(v)} v + y(dv/dy) = (v2logv - 1)/(vlogv)
y(dv/dy) = [(v2logv - 1)/(vlogv)] - v
y(dv/dy) = (v2logv - 1 - v2logv)/vlogv
y(dv/dy) = -(1/vogv)
vlogvdv = -(dy/y)
On integrating both sides,
∫vlogvdv = -∫(dy/y)
logv∫vdv - ∫{d/dv(logv)∫vdv}dv}dv = -∫(dy/y)
(v2/2)logv - (1/2)∫(1/v)(v2/2)dv = -logy + logc
(v2/2)logv - (1/2)∫vdv = -logy + logc
(v2/2)logv - (v2/4) + logy = log|c|
(v2/2)[logv - 1/2] + logy = log|c|
v2[logv - (1/2)] + logy = log|c|
(x2/y2)[log(x/y) - (1/2)] + logy = log|c| (Where ‘c’ is integration constant)
Question 25. (1 + ex/y)dx + ex/y(1 - x/y)dy = 0
Solution:
We have,
(1 + ex/y)dx + ex/y(1 - x/y)dy = 0
\frac{dx}{dy}=-\frac{e^\frac{x}{y}(1-\frac{x}{y})}{1+e^\frac{x}{y}} It is a homogeneous equation,
So, put x = vy (i)
On differentiating both sides w.r.t y,
dx/dy = v + y(dv/dy)
So,
v+y\frac{dv}{dy}=-\frac{e^v(1-v)}{1+e^v} y(dv/dy) = -[ev(1 - v)/(1 + ev)] - v
y(dv/dy) = (-ev + vev - v - vev)/(1 + ev)
y(dv/dy) = -(v + vev)/(1 + ev)
[(1 + ev)/(v + ev)]dv = -(dy/y)
On integrating both sides,
∫[(1 + ev)/(v + ev)]dv = -∫(dy/y)
log|(v + ev)| = -log(y) + log(c)
log|(v + ev)| = log|c/y|
(x/y) + ex/y = c/y
x + yex/y = c (Where ‘c’ is integration constant)
Question 26. (x2 + y2)dy/dx = (8x2 - 3xy + 2y2)
Solution:
We have,
(x2 + y2)dy/dx = (8x2 - 3xy + 2y2)
(dy/dx) = (8x2 - 3xy + 2y2)/(x2 + y2)
It is a homogeneous equation,
So, put y = vx (i)
On differentiating both sides w.r.t x,
dy/dx = v + x(dv/dx)
So,
v + x(dv/dx) = (8x2 - 3xvx + 2v2x2)/(x2 + v2x2)
v + x(dv/dx) = (8 - 3v + 2v2)/(1 + v2)
x(dv/dx) = [(8 - 3v + 2v2)/(1 + v2)] - v
x(dv/dx) = (8 - 4v + 2v2 - v3)/(1 + v2)
(1 + v2)dv/(8 - 4v + 2v2 - v3) = (dx/x)
On integrating both sides,
∫\frac{1+v^2}{4(2-v)+v^2(2-v)}dv=∫\frac{dx}{x}
∫\frac{1+v^2}{(2-v)(4+v^2)}dv=∫\frac{dx}{x} Using partial fraction,
∫\frac{1+v^2}{(2-v)(4+v^2)}=\frac{Av+B}{4+v^2}+\frac{C}{2-v} (1 + v2) = Av(2 - v) + B(2 - v) + C(4 + v2)
(1 + v2) = 2Av - Av2 + 2B - Bv + 4C + Cv2
(1 + v2) = (C - A)v2 + (2A - B)v + (2B + 4C)
Comparing the co-efficient of both sides,
(C - A) = 1
(2A - B) = 0
(2B + 4C) = 1
Solving above equations,
A = -(3/8)
B = -(3/4)
C = (5/8)
∫\frac{-\frac{3}{8}v-\frac{3}{4}}{4+v^2}dv+∫\frac{\frac{5}{8}}{2-v}dv=∫\frac{dx}{x}
-\frac{3}{8}∫\frac{vdv}{v^2+1}-\frac{3}{4}∫\frac{dv}{v^2+1}+\frac{5}{8}∫\frac{dv}{2-v}=logx+logc
-\frac{3}{16}log|v^2+1|-\frac{3}{8}tan^{-1}(\frac{v}{2})-\frac{5}{8}log|2-v|=log|xc|
e^{-\frac{3}{8}tan^{-1}(\frac{v}{2})}=c|x(2-v)^\frac{5}{8}(v^2+4)^\frac{3}{16}|
e^{-\frac{3}{8}tan^{-1}(\frac{y}{2x})}=c|x(2-\frac{y}{x})^\frac{5}{8}(\frac{y^2}{x^2}+4)^\frac{3}{16}|
e^{-\frac{3}{8}tan^{-1}(\frac{y}{2x})}=c|x(2x-y)^\frac{5}{8}(y^2+4x^2)^\frac{3}{16}| (Where ‘c’ is integration constant)
Summary
Exercise 22.9 in Chapter 22 typically focuses on solving first-order linear differential equations. These equations are of the form:
dy/dx + P(x)y = Q(x)
where P(x) and Q(x) are functions of x, and y is a function of x.
Key points:
1. The general solution of a first-order linear differential equation is given by the formula:
y = e^(-∫P(x)dx) [∫Q(x)e^(∫P(x)dx)dx + C]
2. This solution method is known as the integrating factor method.
3. The integrating factor is e^(∫P(x)dx).
4. These equations are widely used in modeling various physical phenomena.
Practice Questions
1. Solve the differential equation: dy/dx + 2xy = x
2. Find the general solution of: dy/dx - y = e^x
3. Solve: (1 + x^2)dy/dx + 2xy = 1
4. Find a particular solution of: dy/dx + y/x = x^2, given that y = 1 when x = 1
5. Solve the initial value problem: dy/dx + y = sin(x), y(0) = 2
6. Find the general solution of: dy/dx + tan(x)y = sec(x)
7. Solve: x dy/dx + 2y = x^3
8. Find the particular solution of: dy/dx + y = e^(-x), given that y = 0 when x = 0
9. Solve the differential equation: dy/dx + y/x = x^n, where n is a constant
10. Find the general solution of: dy/dx + 2y = 4x^2 + 2x + 1
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Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.7 | Set 2
Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.8 | Set 2