6 - Differential Equations Questions Answers
Prove that the system of confocal conics x²/(a² + π) + y²/(b² + π) = 1, π being the parameter is self orthogonal.
let us consider the case of ellipse with x and y axes as their axes
eq. is x2/a2 + y2/b2= 1
on differentiating we get 2x/a2 + 2yy'/b2 = 0 y'is first differential
so yy'/x = -b2/a2
again differentiate and get answer.
You should remember that you should differentiate as many times as the number of constants.
for ex. in the case of parabola only first diffrentiation is sufficient.
now complete it for all conics
using the transformation x=r cosθ and y=r sinθ,find the singular solutionof the differential equation x+py=(x-y)(p2+1)½ where p=dy/dx
x=rcosθ and y=rsinθ
so dx=-rsinθdθ and dy = rcosθdθ
so p = -cotθ
so given eq. will become x-cotθy = (x-y)cosecθ
on putting values of x and y we get
0 = r(cosθ-sinθ)/sinθ
so tanθ = 1 so θ = nπ+π/4
so x = rcos(nπ+π/4) and y = rsin(nπ+π/4)
show that the family of parabolas y2=4a(x+a) is self orthogonal.
y2= 4a(x+a)
so 2yy' = 4a so a = yy'/2
on putting a we get y2= 4yy'/2(x+yy'/2)
so y2 = yy' (2x+yy') or y = 2xy' + yy'2 (1)
now on putting -1/y' in the place of y'
we get y2 = -y/y'[2x-y/y']
so -yy'2 = 2xy' - y (2)
similarity of (1) and (2) shows that the given curve is self orthogonal
how to find differential eqn. of all conics whose axes coincide with coordinate axes?? tell me the eqn. of tht. conic
Parabola eq. y2 = 4ax ,
on differentiating this eq. with respect to x we get 2y(dy/dx) = 4a
on taking this value of 4a in eq. y2= 4ax we will get the differential eq. of parabola.
similarly for ellipse x2/a2 + y2/b2 = 1
on differentiaing this eq. twice we will get two additional eq.
by solving these eq. get an eq. free from a and b, this will be the differential eq. of the ellipse.
similarly for hyperbola
Solve: [3Öxy + 4y - 7Öy]dx + [4x - 7Öxy + 5Öx]dy = 0.
[3Öxy + 4y - 7Öy]dx + [4x - 7Öxy + 5Öx]dy = 0
=> dy/dx = [3Öxy + 4y - 7Öy] / [-4x + 7Öxy - 5Öx]
=> dy/dx = √y/√x [(3√x + 4√y - 7) / (-4√x + 7√y - 5)]
=> (dy/√y)/(dx/√x) = [(3√x + 4√y - 7) / (-4√x + 7√y - 5)]
now take √x = X + h and √y = Y + k
you will get a homogeneous equation dY/dX = (3X+4Y+3h+4k-7)/(7Y-4X+7k-4h-5)
take 3h+4k-7 = 0 and 7k-4h-5 = 0 and solve the equation
[3Öxy + 4y - 7Öy]dx + [4x - 7Öxy + 5Öx]dy = 0
=> dy/dx = [3Öxy + 4y - 7Öy] / [-4x + 7Öxy - 5Öx]
=> dy/dx = √y/√x [(3√x + 4√y - 7) / (-4√x + 7√y - 5)]
=> (dy/√y)/(dx/√x) = [(3√x + 4√y - 7) / (-4√x + 7√y - 5)]
now take √x = X + h and √y = Y + k
you will get a homogeneous equation dY/dX = (3X+4Y+3h+4k-7)/(7Y-4X+7k-4h-5)
take 3h+4k-7 = 0 and 7k-4h-5 = 0 and solve the equation
dY/dX = (3X+4Y)/(7Y-4X)
take Y = uX then dY/dX = u + Xdu/dX
after solving put the values of X and Y in terms of x and y,