The solution of 1 + x2dydx + 2xy - 4x2 = 0 is :
3x1 + y2 = 4y3 + c
3y(1 + x2) = 4x3 + c
3x(1 + y2) = 4y3 + c
3y(1 + y2) = 4x3 + c
The solution of dxdy + xy = x2 is :
1y = cx - xlogx
1x = cy - ylogy
1x = cx + xlogy
1y = cx - ylogx
B.
dxdy + xy = x2⇒ 1x2dxdy + 1xy = 1Put 1x = t ⇒ - 1x2dxdy = dtdy∴ - dtdy + ty = 1 ⇒ dtdy - ty = - 1On compairing with dydx + Py = Q, we getP = - 1y, Q = - 1I.F. = e∫pdx = e- ∫1ydy = 1yThe solution ist(I.F.) = ∫QI.F.dy + c t . 1y = ∫- 1 . 1ydy + c 1x1y = - logy + c 1x = cy - y logy
The differential equation obtained by eliminating the arbitrary constants a and b from xy = aex + be- x is
xd2ydx2 + 2dydx - xy = 0
xd2ydx2 + 2ydydx - xy = 0
xd2ydx2 + 2dydx + xy = 0
d2ydx2 + dydx - xy = 0
The solution of x + y + 1dydx = 1 is
y = (x + 2) + cex
y = - (x + 2) + cex
x = - (y + 2) + cey
x = (y + 2)2 + cey
The solution of dydx = y2xy - x2 is
eyx = kx
eyx = ky
exy = kx
e - yx = ky
The solution of dydx + 1 = ex + y is
e - x + y + x + c = 0
e - x + y - x + c = 0
e x + y + x + c = 0
e x + y - x + c = 0
The solution of the differential equation
dydx = xy + yxy + x is
x + y = logcyx
x + y = logcxy
x - y - logcxy
y - x = logcxy
dydx = x - 2y + 12x - 4y is
(x - 2y)2 + 2x = c
(x - 2y)2 + x = c
(x - 2y)2 + 2x2 = c
(x - 2y) + x2 = c
The solution of the differential equation dydx - ytanx = exsecx is
y = excosx + c
ycosx = ex + c
y = exsinx + c
ysinx = ex + c
xy2dy - x3 + y3dx = 0 is
y3 = 3x3 + c
y3 = 3x3 logcx
y3 = 3x3 + logcx
y3 + 3x3 = logcx