The differential equation of all straight lines passing through the point (1, - 1), is
y = x + 1dydx + 1
y = x + 1dydx - 1
y = x - 1dydx + 1
y = x - 1dydx - 1
Integrating factor of differential equation cosxdydx + ysinx = 1 is
cosx
tanx
secx
sinx
The order and degree of the differential equation d2ydx2 + dydx13 + x14 = 0 are respectively
2, 3
3, 3
2, 6
2, 4
The general solution of the differential equation ydx + (1 + x2) tan- 1(x)dy = 0, is
ytan-1x = c
xtan-1y = c
y + tan-1x = c
x + tan-1y = c
Solution of differential equation dydx = 2xy is
y = cex2
y2 = 2x2 + c
y = ce- x2
y = x2 + c
The second order differential equation is
y' + x = y2
y'y'' + y = sin(x)
y''' + y'' + y = 0
y' = y
The solution of the differential equation dydx - yx = 1 is
x2loge(x) + y = c
xloge(x) + cx = y
x2loge(x) - y = c
xloge(x) + y = cx
B.
The given differential equation isdydx - yx = 1which is a linear differential equation.Here, P = - 1x and Q = 1∴ IF = e∫Pdx = e- ∫1xdx = elogx- 1 = 1xHence, required solution is yIF = ∫IFQdx + c⇒ y1x = ∫1x . 1dx + c⇒ y = xlogex + cx
An integrating factor of the differential equation 1 - x2dydx - xy = 1, is
- x
x1 - x2
1 - x2
12log1 - x2
Solve dydxtany = sinx + y + sinx - y
secx - 12tany = c
logsinx + y = c
secx + tany = c
secy + 2cosx = c
Find the differential equation of curves y = Aex + Be-x fordifferent values of A and B
d2ydx2 - 2y = 0
d2ydx2 = y
d2ydx2 = 4y + 3
d2ydx2 + y = 0