The solution of the differential equation dydx = sinx + ytanx + y - 1 is
cscx + y + tanx + y = x + c
x + cscx + y = c
x + tanx + y = c
x + secx + y = c
The differential equation of the family y = aex + bxex + cx2ex of curves, where a, b, c are arbitrary constants , is
y''' + 3y'' + 3' + y = 0
y''' + 3y'' - 3' - y = 0
y''' - 3y'' - 3' + y = 0
y''' - 3y'' + 3' - y = 0
The solution of tanydydx = sinx + y + sinx - y is
sec(y) = 2cos(x) + c
sec(y) = - 2cos(x) + c
tan(y) = - 2cos(x) + c
sec2(y) = - 2cos(x) + c
A family of curves has the differential equation xydydx = 2y2 - x2. Then, the family of curves is
y2 = cx2 + x3
y2 = cx4 + x3
y2 = x + cx4
y2 = x2 + cx4
The solution of the differential equation dydx = yx + ϕyxϕ'yx is
xϕyx = k
ϕyx = kx
yϕyx = k
ϕyx = ky
If y = y(x) is the solution of the differential equation 2 + sinxy + 1dydx + cosx = 0 then y(π2)is equal to
13
23
1
43
If u = fr, where r2 = x2 + y2, then∂u∂x2 + ∂2u∂y2 = ?
f''(r)
f''(r) +f'(r)
f''(r) + 1rf'(r)
f''(r) + rf'(r)
If dydx + 2xtanx - y = 1, then sinx - y = ?
Ae- x2
Ae2x
Aex2
Ae - 2x
C.
Given differential equation isdydx + 2xtanx - y = 1Put x - y =t⇒ 1 - dydx = dtdx⇒ dydx =1 - dtdx∴ 1 - dtdx + 2xtant = 1⇒ dttant = 2xdx⇒ cottdt = 2xdxOn integrating both sides, we getlogsint = x2 + logA⇒ logsinx - y A = x2⇒ sinx - y = Aex2
An integrating factor of the differential equation1 - x2dydx + xy = x41 + x51 - x23 is
1 - x2
x1 - x2
x21 - x2
11 - x2
If cos-1yb = 2logx2, where x > 0, thenx2d2ydx2 + xdydx = ?
4y
- 4y
0
- 8y