The solution of the differential equation dydx = y tanx - 2 sinx, is
y sinx = c + sin2x
y cosx = c + 12sin2x
y cosx = c - sin2x
y cosx = c + 12cos2x
The differential equation of system of concentric circles with centre (1, 2) is :
x - 2 + y - 1dydx = 0
x - 1 + y - 2dydx = 0
x + 1dydx + y - 2 = 0
x + 2dydx + y - 1 = 0
The solution of the differential equation dydx + 2yx1 + x2 = 11 + x22 is :
y(1 + x2) = c + tan-1(x)
ylog1 + x2 = c + tan-1x
y1 + x2 = c + tan-1x
y1 + x2 = c + sin-1x
The solution of the differential equation xdy - ydx = x2 + y2dx is :
x+ x2 + y2 = cx2
y- x2 + y2 = cx
x - x2 + y2 = cx
y+ x2 + y2 = cx2
The solution of the differential equation dydx = ex - y + x2e- y is :
y = ex - y + x2e- y + c
ey - ex = 13x3 + c
ey + ex = 13x3 + c
ex - ey = 13x3 + c
The integrating factor of the differential equation dydx + 1xy = 3x is :
x
in x
0
∞
The solution of the differential equation sec2(x)tan(y))dx + sec2(y)tan(x))dy = 0 is :
tanytanx = c
tan2x tany= c
None of these
The differential equation of all straight lines passing through origin is :
y = xdydx
dydx = y + x
dydx = y - x
To reduce the differential equation dydx = Py = Qx . yn to the linear form, the substitution is :
v = 1yn
v = 1yn - 1
v = yn
v = yn - 1
B.
Given, dydx = Py = Qx . yn⇒ 1yn . dydx + y-n + 1 Px = QxSubstituting 1yn - 1 = v ...i⇒ dvdx = - n + 1y- ndydx∴ dvdx + 1 - n Px . v = 1 - n QxThis is a linear differential equation of the formdydx + Py = Q∴ From Eq. (i) proper subtitution is v = 1yn - 1
Integrating factor of the differential equation dydx + Pxy = Qx is :
∫P dx
∫Q dx
e∫P dx
e∫Q dx