# 12 - Differential Equations Questions Answers

show that the family of the parabolas y^2=2cx+c^2 is self - orthogonal

**Asked By: MAGDY MOHMMAD**1 year ago

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show that the family of the parabolas y^2=2cx+c^2 is self - orthogonal

**Asked By: MAGDY MOHMMAD**1 year ago

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What is the solution of Differential Equation y=k(x-k)^{2}?

**Asked By: VIJAYENDRA KUMAR**4 year ago

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**Asked By: MONARK SUTARIYA**5 year ago

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**Asked By: NEHA GUPTA**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

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

**Asked By: NEHA GUPTA**7 year ago

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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**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

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**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

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

Solve: dy/dx =[x√(x^2-1) +y]/(√x^2-1)

**Asked By: SUBHADEEP BASU**7 year ago

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