The general solution of the equation is
y2 = (1 + x)log(1 + x) - c
y2 = 1 - xlogc1 - x - 1
y2 = 1 + xlogc1 + x - 1
D.
dydx = y2 - x2yx + 1 ...i⇒ ydydx = y2 - x2x + 1⇒ ydydx - y22x + 1 = - x2x + 1 ...iiPut y2 = tBy differentiating both side w. r. t. 'x', we get 2ydydx = dtdx∴ Eq. (ii) reduces to12dtdx - t2x + 1 = - x2x + 1This is linear differential equation withP = - 1x + 1 and Q = - xx + 1
∴ IF = e∫Pdx = e∫- 1x + 1dx = e- logx + 1 = elog1x + 1 = 1x + 1∴Required solution will bet IF = ∫QIFdx + logc Here, logc is a constanty2 . 11 + x = ∫- xx + 1 × 1x + 1dx + logc⇒ y21 + x = - ∫x + 1 - 1x + 12dx + logc⇒ y21 + x = - ∫11 + xdx - ∫11 + x2dx + logc⇒ y21 + x = - log1 + x + 11 + x + logc⇒ y21 + x = logc1 + x - 11 + x⇒ y2 = 1 + xlogc1 + x - 1
The general solution of the differential equation dydx + sinx + y2 = sinx - y2 is
logetany2 = - 2sinx2 + C
logetany4 = 2sinx2 + C
logetany4 = - 2sinx2 + C
If a, b, c are three vectors such that [a b c] = 5, then the value of [a x b, b x c, c x a] is
15
25
20
10
The point of inflection of the function y = ∫0xt2 - 3t + 2dt is
32, 34
- 32, - 34
- 12, - 32
12, 32
The value of integral ∫dxxx2 - a2
c - 1asin-1ax
c - 1acos-1ax
sin-1ax + c
c + 1asin-1ax
The function y specified implicitly by the relation ∫0yetdt + ∫0xcostdt = 0 satisfies the differential equation
e2yd2ydx2 + dydx2 = sinx
eyd2ydx2 + dydx2 = sin2x
ey2d2ydx2 + dydx2 = sinx
eyd2ydx2 + dydx2 = sinx