# 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.

**Asked By: SEVAK KUMAR**

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**Asked By: NEHA GUPTA**

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**Joshi sir comment**

let us consider the case of ellipse with x and y axes as their axes

eq. is x^{2}/a^{2} + y^{2}/b^{2}= 1

on differentiating we get 2x/a^{2} + 2yy^{'}/b^{2 }= 0 y'is first differential

so yy^{'}/x = -b^{2}/a^{2}

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)(p^{2}+1)½ where p=dy/dx

**Asked By: DEBANJAN GHOSHAL**

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**Joshi sir comment**

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 y^{2}=4a(x+a) is self orthogonal.

**Asked By: DEBANJAN GHOSHAL**

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**Joshi sir comment**

y^{2}= 4a(x+a)

so 2yy^{'} = 4a so a = yy^{'}/2

on putting a we get y^{2}= 4yy^{'}/2(x+yy^{'}/2)

so y^{2} = yy^{'} (2x+yy^{'}) or y = 2xy^{'} + yy^{'2} (1)

now on putting -1/y^{'} in the place of y^{'}

we get y^{2} = -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

**Asked By: SHALINI SINGH**

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**Joshi sir comment**

Parabola eq. y^{2} = 4ax ,

on differentiating this eq. with respect to x we get 2y(dy/dx) = 4a

on taking this value of 4a in eq. y^{2}= 4ax we will get the differential eq. of parabola.

similarly for ellipse x^{2}/a^{2} + y^{2}/b^{2} = 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.

**Asked By: KAMAL**

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**Joshi sir comment**

[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

**Solution by Joshi sir**

[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,