Question 14. sin4x(dy/dx) = cosx
Solution:
We have,
sin4x(dy/dx) = cosx
dy = (cosx/sin4x)dx
Let, sinx = z
On differentiating both sides, we get
cosx dx = dz
dy = (dz/z4)
On integrating both sides, we get
∫(dy) = ∫(1/z4)dz
y = (1/ -3t3) + c
y = -(1/3sin3x) + c
y = (-cosec3x/3) + c -(Here, 'c' is integration constant)
Question 15. cosx(dy/dx) - cos2x = cos3x
Solution:
We have,
cosx(dy/dx) - cos2x = cos3x
(dy/dx) = (cos3x + cos2x)/cosx
(dy/dx) = (4cos3x - 3cosx + 2cos2x - 1)/cosx
(dy/dx) = (4cos3x/cosx) - 3(cosx/cosx) + 2(cos2x/cosx) - secx
dy = [4cos2x - 3 + 2cosx - secx]dx
dy = [4{(cos2x + 1)/2} - 3 + 2cosx - secx]dx
On integrating both sides, we get
∫dy = ∫[2cos2x - 1 + 2cosx - secx]dx
y = sin2x - x + 2sinx - log|secx + tanx| + c -(Here, 'c' is integration constant)
Question 16. √(1 - x4)(dy/dx) = xdx
Solution:
We have,
√(1 - x4)(dy/dx) = xdx
Let, x2 = z
On differentiating both sides, we get
2xdx = dz
xdx = (dz/2)
√(1 - z2)dy = (dz/2)
dy =
\frac{dz}{2\sqrt{1-z^2}} On integrating both sides, we get
∫dy = ∫
\frac{dz}{2\sqrt{1-z^2}} y = (1/2)sin-1(z) + c
y = (1/2)sin-1(x2) + c -(Here, 'c' is integration constant)
Question 17. √(a + x)(dy) + xdx = 0
Solution:
We have,
√(a + x)(dy) + xdx = 0
dy =
\frac{-x}{\sqrt{a + x}} dxLet, (x + a) = z2
On differentiating both sides, we get
dx = 2zdz
(x + a) = z2
x = z2 - a
dy = -2[(z2 - a)/z]zdz
On integrating both sides, we get
∫dy = -2∫[(z2 - a)/z]zdz
y = -(2/3)(z3) + 2az + c
y = -(2/3)(x + a)3/2 + 2a√(x + a) + c -(Here, 'c' is integration constant)
Question 18. (1 + x2)(dy/dx) - x = 2tan-1x
Solution:
We have,
(1 + x2)(dy/dx) - x = 2tan-1x
(1 + x2)(dy/dx) = 2tan-1x + x
dy/dx =
\frac{2tan^{-1}x + x}{1 + x^2} dy
=(\frac{2tan^{-1}x+x}{1+x^2})dx On integrating both sides, we get
∫dy=∫(\frac{2tan^{-1}x+x}{1+x^2})dx y = I1 + I2
I1 = ∫(
\frac{2tan^{-1}x}{1 + x^2} Let, tan-1x = z
On differentiating both sides, we get
\frac{dx}{1 + x^2} = dz= ∫2zdx
= z2
I1 = (tan-1x)2
I2 = ∫
\frac{xdx}{(1+x^2)} = (1/2)log|1 + x2|
y = (tan-1x)2 + 1/2log|1 + x2|+ c -(Here, 'c' is integration constant)
Question 19. (dy/dx) = xlogx
Solution:
We have,
(dy/dx) = xlogx
dy = xlogxdx
On integrating both sides, we get
∫dy = ∫xlogxdx
y = log|x|∫xdx - ∫[
\frac{(logx)}{dx} ∫xdx]dxy = (x2/2)log|x| - ∫(1/x)(x2/2)dx
y = (x2/2)log|x| - ∫(x/2)dx
y = (x2/2)log|x| - (x2/4) + c -(Here, 'c' is integration constant)
Question 20. (dy/dx) = xex - (5/2) + cos2x
Solution:
We have,
(dy/dx) = xex - (5/2) + cos2x
dy = (xex - (5/2) + cos2x) dx
On integrating both sides, we get
∫dy = ∫xex dx - 5/2∫dx + ∫cos2x dx
y = ∫xex dx - 5/2∫dx + ∫(1 + cos2x)/2 dx
= ∫xex dx - 5/2∫dx + 1/2∫dx + 1/2∫cos2x dx
= ∫xex dx - 2∫dx + 1/2∫cos2x dx
= x∫ex dx - ∫(1∫ex dx)dx - 2x + sin2x/4 dx
= xex - ex - 2x + 1/4sin2x + c
Question 21. (x3 + x2 + x + 1)(dy/dx) = 2x2 + x
Solution:
We have,
(x3 + x2 + x + 1)(dy/dx) = 2x2 + x
(dy/dx) = (2x2 + x)/(x3 + x2 + x + 1)
dy =
\frac{2x^2+x}{(x+1)(x^2+1)}dx On integrating both sides, we get
∫dy = ∫
\frac{2x^2+x}{(x+1)(x^2+1)}dx Let,
\frac{2x^2+x}{(x+1)(x^2+1)}dx=\frac{A}{(x+1)}+\frac{Bx+C}{x^2+1} 2x2 + x = Ax2 + A + Bx2 + Bx + Cx + C
2x2 + x = (A + B)x2 + (B + C)x + (A + C)
On comparing the coefficients on both sides,
(A + B) = 2
(B + C) = 1
(A + C) = 0
After solving the equations,
A = (1/2)
B = (3/2)
C = -(1/2)
y = (1/2)∫(dx/(x + 1) +
∫\frac{\frac{3}{2}x-\frac{1}{2}}{x^2+1}dx y = (1/2)log(x + 1) + (3/4)∫
\frac{2x}{x^2 + 1} dx - (1/2)∫\frac{dx}{(x^2 + 1)} y = (1/2)log|x + 1| + (3/4)log|x2 + 1| - (1/2)tan-1x + c -(Here, 'c' is integration constant)
Question 22. sin(dy/dx) = k, y(0) = 1
Solution:
We have,
sin(dy/dx) = k,
(dy/dx) = sin-1(k)
dy = sin-1(k)dx
On integrating both sides, we get
∫dy = sin-1(k)∫dx
y = xsin-1k + c -(1)
Put x = 0, y = 1
1 = 0 + c
1 = c
On putting the value of c in equation(1)
y = xsin-1k + 1
y - 1 = xsin-1x
Question 23. e(dy/dx) = x + 1, y(0) = 3
Solution:
We have,
e(dy/dx) = x + 1
(dy/dx) = log(x + 1)
dy = log(x + 1)dx
On integrating both sides, we get
∫dy = ∫log(x + 1)dx
y = log(x + 1)∫dx - ∫[
\frac{dlog(x+1)}{dx} ∫dx]dxy = xlog(x + 1) - ∫[x/(x + 1)]dx
y = xlog(x + 1) - ∫[
1 - \frac{1}{(x+1)} ]dxy = xlog(x + 1) - x + log(x + 1) + c
y = (x + 1)log(x + 1) - x + c -(1)
Put, y = 3, x = 0 in equation(1)
3 = 0 + c
y = (x + 1)log(x + 1) - x + 3
Question 24. c'(x) = 2 + 0.15x, c(0) = 100
Solution:
We have,
c'(x) = 2 + 0.15x -(1)
On integrating both sides, we get
∫c'(x)dx = ∫(2 + 0.15x)dx
c(x) = 2x + 0.15(x2/2) + c -(2)
Put, c(0) = 100, x = 0 in equation(2)
100 = 2(0) + 0 + c
c = 100
c(x) = 2x + 0.15(x2/2) + 100
Question 25. x(dy/dx) + 1 = 0, y(-1) = 0
Solution:
We have,
x(dy/dx) + 1 = 0
xdy = -dx
dy = -(dx/x)
On integrating both sides, we get
∫dy = -∫(dx/x)
y = -logx + c -(1)
Put, y = 0, x = -1 in equation(1)
0 = 0 + c
c = 0
y = -log|x|
Question 26. x(x2 - 1)(dy/dx) = 1, y(2) = 0
Solution:
We have,
x(x2 - 1)(dy/dx) = 1 -(1)
dy = dx/x(x + 1)(x - 1)
On integrating both sides, we get
∫dy = ∫dx/x(x + 1)(x - 1)
Let, 1/x(x + 1)(x - 1) = A/x + B/(x + 1) + C/(x - 1)
1 = A(x + 1)(x - 1) + B(x)(x - 1) + C(x)(x + 1) -(2)
Put, x = 0, -1, 1 respectively and simplify above equation, we get,
A = -1, B = (1/2), C = (1/2)
y =
-∫\frac{dx}{x}+∫\frac{1}{2}\frac{dx}{(x+1)}+∫\frac{1}{2}\frac{dx}{(x-1)} y = -logx + (1/2)log(x + 1) + (1/2)log(x - 1)
y = (1/2)log(1/x2) + (1/2)log(x + 1) + (1/2)log(x - 1) + c -(3)
Put, y = 0, x = 2 in equation(3)
0 = (1/2)log(1/4) + (1/2)log(3) + 0 + c
c = -(1/2)log(3/4)
y = (1/2)log[(x2 - 1)/x2] - (1/2)log(3/4)
y=\frac{1}{2}log[\frac{4}{3}\frac{(x^2-1)}{x^2}]
Summary
Exercise 22.5, Set 2 focuses on solving separable differential equations. These equations can be written in the form dy/dx = f(x)/g(y), where f(x) is a function of x only and g(y) is a function of y only. The problems in this set involve a variety of functions including polynomial, exponential, trigonometric, and logarithmic functions. Students are asked to find both general and particular solutions. The primary method used to solve these equations involves separating the variables and then integrating both sides. This exercise helps students recognize separable forms, apply the separation of variables technique, and solve the resulting integrals.
