The solution of is
y = ce- ∫Pdx
x = ce- ∫Pdy
x = ce∫Pdy
B.
∵ In the given linear differential equation coefficient of Q is 0.
∴ Solution isy . e∫P dx = ∫0 dx + c⇒ y = ce- ∫P dx
The differential equation of the family of lines passing through the origin is :
xdydx + y = 0
x + dydx = 0
dydx = y
xdydx - y = 0
The solution of dydx + y = e- x; y(0) = 0 is :
y = e- x(x - 1)
y = xe- x
y = xe- x + 1
y = (x + 1)e-x
∫- 221 - x2dx is equal to :
4
2
- 2
0
∫tanxsinxcosxdx is equal to :
2tanx + c
cotx + c
tan2x + c
∫dxx2 + 4x + 13 is equal to :
logx2 + 4x + 130 + c
13tan-1x + 23 + c
log2x + 4 + c
1x2 + 4x + 13 + c
∫01ddxsin-12x1 + x2dx is equal to :
π
π2
π4
∫0π2sinxsinx + cosxdx is equal to
π3
The area bounded by the parabolas y2 = 4ax and x2 = 4ay is :
8a33 sq unit
16a23 sq unit
32a23 sq unit
64a23 sq unit
The area of the regior x, y : x2 + y2 ≤ 1 ≤ x + y is :
π25 sq unit
π22 sq unit
π23 sq unit
π4 - 12 sq unit