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Solution:
We have,
(1 - x2)dy + xydx = xy2dx
(1 - x2)dy = xy2dx - xydx
(1 - x2)dy = xy(y - 1)dx
On integrating both sides,
log(y - 1) - logy = -(1/2)log(1 - x2) + logc
log(y - 1) - logy + (1/2)log(1 - x2) = logc (Where 'c' is integration constant)
Solution:
We have,
tanydx + sec2ytanxdy = 0
tanydx = -sec2ytanxdy
(sec2y/tany)dy = -dx/tanx
On integrating both sides,
∫(sec2y/tany)dy = -∫cotxdx
Let, tany = z
On differentiating both sides
sec2xdx = dz
∫(dz/z) = -∫cotxdx
log(z) = -log(sinx) + log(c)
On putting the value of z in above equation
log(tany) + log(sinx) = log(c)
log[(sinx)(tany)] = log(c)
sinx.tany = c (Where 'c' is integration constant)
Solution:
We have,
(1 + x)(1 + y2)dx + (1 + y)(1 +x2)dy = 0
On integrating both sides,
tan-1(y) + (1/2)log(1 + y2) = -tan-1(x) - (1/2)log(1 + x2) + c
tan-1(y) + tan-1(x) + (1/2)log[(1 + y2)(1 + x2)] = c (Where 'c' is integration constant)
Solution:
We have,
tany(dy/dx) = sin(x + y) + sin(x - y)
tany(dy/dx) = 2sin{(x + y + x - y)/2}cos{(x + y - x + y)/2}
tany(dy/dx) = 2sinxcosy
(tany/cosy)dy = 2sinxdx
On integrating both sides,
∫secytanydy = 2∫sinxdx
secy = -2cosx + c
secy + cosx = c (Where 'c' is integration constant)
Solution:
We have,
cosxcosy(dy/dx) = -sinxsiny
(cosy/siny)dy = -(sinx/cosx)dx
cotydy = -tanxdx
On integrating both sides,
∫cotydy = -∫tanxdx
log(siny) = log(cosx) + logc
log(siny) = log(cosx.c)
siny = c.cosx (Where 'c' is integration constant)
Solution:
We have,
(dy/dx) + cosxsiny/cosy = 0
(dy/dx) = -cosx.tany
dy/tany = -cosxdx
cotydy = -cosxdx
On integrating both sides,
∫cotydy = -∫cosxdx
log(cosy) = -sinx + c
log(cosy) + sinx = c (Where 'c' is integration constant)
Solution:
We have,
x√(1 - y2)(dx) + y√(1 - x2)dy = 0
x√(1 - y2)(dx) = -y√(1 - x2)dy
On integrating both sides,
√(1 - y2) = -√(1 - x2) + c
√(1 - y2) + √(1 - x2) = c (Where 'c' is integration constant)
Solution:
We have,
y(1 + ex)dy =(y + 1)exdx
On integrating both sides,
∫[1 - 1/(y + 1)]dy = ∫exdx/(1 + ex)
y - log(y + 1) = log(1 + ex) + c (Where 'c' is integration constant)
Solution:
We have,
(y + xy)dx + (x - xy2)dy = 0
y(1 + x)dx = -x(1 - y2)dy
[(1 - y2)/y]dy = -[(1 + x)/x]dx
On integrating both sides,
∫[(1 - y2)/y]dy = -∫[(1 + x)/x]dx
∫(dy/y) - ∫ydy = -∫dx/x - ∫dx
log(y) - (y2/2) = -log(x) - x + c
log(x) + x + log(y) - (y2/2) = c (Where 'c' is integration constant)
Solution:
We have,
(dy/dx) = 1 - x + y - xy
(dy/dx) = (1 - x) + y(1 - x)
(dy/dx) = (1 - x)(1 - y)
dy/(1 - y) = (1 - x)dx
On integrating both sides,
∫dy/(1 - y) = ∫(1 - x)dx
log(1 - y) = x - (x2/2) + c (Where 'c' is integration constant)
Solution:
We have,
(y2 + 1)dx - (x2 + 1)dy = 0
(y2 + 1)dx = (x2 + 1)dy
On integrating both sides,
tan-1y = tan-1x + c (Where 'c' is integration constant)
Solution:
We have,
dy + (x + 1)(y + 1)dx = 0
dy/(y + 1) = -(x + 1)dx
On integrating both sides,
∫dy/(y + 1) = -∫(x + 1)dx
log(y + 1) = -(x2/2) - x + c
log(y + 1) + (x2/2) + x = c (Where 'c' is integration constant)
Solution:
We have,
(dy/dx) = (1 + x2)(1 + y2)
On integrating both sides,
tan-1y = x + (x3/3) + c
tan-1y - x - (x3/3) = c (Where 'c' is integration constant)
Solution:
We have,
(x - 1)(dy/dx) = 2x3y
dy/y = 2x3dx/(x - 1)
On integrating both sides,
∫dy/y = 2∫x3dx/(x - 1)
log(y) = (2/3)(x3) + 2(x2/2) + 2x + 2log(x - 1) + log(c)
y = c|x - 1|2e[(2/3)x3+x2+2x] (Where 'c' is integration constant)
Solution:
We have,
(dy/dx) = ex+y + e-x+y
(dy/dx) = ex.ey + e-x.ey
(dy/dx) = ey(ex + e-x)
dy/ey = (ex + e-x)dx
On integrating both sides,
∫e-ydy = ∫exdx + ∫e-xdx
-e-y = ex - e-x + c
e-x-e-y = ex + c (Where 'c' is integration constant)
Solution:
We have,
(dy/dx) = (cos2x - sin2x)cos2y
dy/cos2y = (cos2x - sin2x)dx
sex2ydy = cos2xdx
On integrating both sides,
∫sex2ydy = ∫cos2xdx
tany = (sin2x/2) + c (Where 'c' is integration constant)
Solution:
We have,
(xy2 + 2x)(dx) + (x2y + 2y)dy = 0
x(y2 + 2)(dx) = -y(x2 + 2)dy
Multiplying both sides by 2,
On integrating both sides,
log(y2 + 1) = -log(x2 + 1) + log(c)
(Where 'c' is integration constant)
Solution:
We have,
cosecx logy(dy/dx) + x2y2 = 0
log(y)dy/y2 = -x2dx/cosecx
On integrating both sides,
∫[log(y)/y2]dy = -∫x2sinxdx
-log(y)/y + ∫dy/y2 = x2cosx - 2∫xcosxdx + c
-log(y)/y - 1/y = x2cosx - 2[x∫cosxdx - ∫{dx/dx∫cosxdx}dx] + c
-[{log(y) + 1}/y] = x2cosx - 2(xsinx - ∫sinxdx) + c
x2cosx + [{log(y) + 1}/y] - 2(xsinx + cosx) = c
Solution:
We have,
xy(dy/dx) = 1 + x + y + xy
xy(dy/dx) = (1 + x) + y(1 + x)
xy(dy/dx) = (1 + x)(1 + y)
ydy/(1 + y) = [(1 + x)/x]dx
On integrating both sides,
∫ydy/(1 + y) = ∫[(1 + x)/x]dx
∫[1 - 1/(1 + y)]dy = ∫(dx/x) + ∫dx
y - log(1 + y) = log(x) + x + log(c)
y = log(x) + log(1 + y) + x + log(c)
y = log[cx(1 + y)] + x (Where 'c' is integration constant)
Solution:
We have,
y(1 - x2)(dy/dx) = x(1 + y2)
On integrating both sides,
Multiplying both sides by 2,
log(1 + y2) = -log(1 - x2) + log(c)
log[(1 + y2)(1 - x2)] = logc
(1 + y2)(1 - x2) = c (Where 'c' is integration constant)
Solution:
We have,
yex/ydx = (xex/y + y2)dy
yex/ydx - xex/ydy = y2dy
ex/y(ydx - xdy)/y2 = dy
ex/yd(x/y) = dy
On integrating both sides,
∫ex/yd(x/y) = ∫dy
ex/y = y + c (Where 'c' is integration constant)
Solution:
We have,
(1 + y2)tan-1xdx + 2y(1 + x2)dy = 0 -(i)
On integrating both sides,
-(ii)
Let, I =
2I = (1/2)(tan-1x)2
I = (1/4)(tan-1x)2
From equation (ii)
(1/2)log(1 + y2) = -(1/4)(tan-1x)2 + c
log(1 + y2) + (1/2)(tan-1x)2 = c
Solution:
We have,
(dy/dx) = ytan2x
(dy/y) = tan2xdx
On integrating both sides,
∫(dy/y) = ∫tan2xdx
log(y) = (1/2)log(sec2x) + log(c)
y = c(sec2x)1/2
Put x = 0, y = 2 in above equation
c = 2
y = 2(sec2x)1/2
Exercise 22.7, Set 2 continues the focus on homogeneous differential equations of the first order. These problems involve equations where the right-hand side can be expressed as a function of y/x. The set includes a mix of general solution problems and particular solution problems with given initial conditions. The complexity ranges from simple rational expressions to more involved quadratic forms. The primary solving technique involves the substitution y = vx to transform the homogeneous equation into a separable one. This exercise reinforces students' ability to recognize homogeneous forms, apply the appropriate substitution, and solve the resulting separable equations.
