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Solved Examples

Question 1: Prove that the envelope of a plane, the sum of the squares of whose intercepts on the coordinate axes is constant, is a surface x2/3 + y2/3 + z2/3 = Constant.
Solution:
Let the equation of the plane be

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c}$ = 1                                    ......................(1)

where a2 + b2 + c2= k2                                            ......................(2)

Eliminating 'c' the equation of the plane becomes

F = $\frac{x}{a} + \frac{y}{b} + \frac{z}{\sqrt{k^2 - a^2 - b^2}} $ - 1 = 0

Therefore $\frac{\partial F}{\partial a}$ = 0

=> $\frac{-x}{a^2}- \frac{1}{2}z(k^2 - a^2 - b^2)^{\frac{-3}{2}}$ (-2a)            .....................(3)

and $\frac{\partial F }{\partial b}$ = 0

=> $\frac{-y}{b^2}- \frac{1}{2}z(k^2 - a^2 - b^2)^{\frac{-3}{2}}$(-2b)            ......................(4)

By equation (3) and equation (4) we have

$\frac{x}{a^2} * \frac{b^2}{y} = \frac{a}{b}$

=> $\frac{x}{a^3}$ = $\frac{y}{b^3}$ = $\frac{z}{c^3}$
 
or $\frac{\frac{x}{a} + \frac{y}{b} + \frac{z}{C}}{a^2 + b^2 + c^2} = \frac{1}{k^2}$         ........................(5)


From equation (5) we have

a = k2/3 x1/3, b = k2/3 y1/3, c = k2/3 z1/3

Put the values of a, b, c in equation (1), the envelope of the plane is

x2/3 + y2/3 + z2/3 = k2/3  = Constant
 

Question 2: Evaluate $\Delta$(Sinx Cos3x)
Solution:
$\Delta$(Sinx Cos3x)

= $\Delta${$\frac{1}{2}$(2Sinx Cos3x)}

or $\Delta${$\frac{1}{2}$(2Cos3x Sinx)}

[2Cos A Sin B = Sin(A + B) - Sin(A - B)]

= $\Delta${$\frac{1}{2}$(Sin 4x - Sin 2x)}

= $\frac{1}{2}${$\Delta$ Sin4x - $\Delta$ Sin2x}

= $\frac{1}{2}${Sin4(x + h) - Sin4x} - {Sin2(x + h) - Sin 2x}

[Sin A - Sin B = 2Cos($\frac{A + B}{2}$) Sin($\frac{A - B}{2}$)]

= $\frac{1}{2}${2Cos(4x + 2h)Sin2h - 2Cos(2x + h)Sinh}

= Cos(4x + 2h)Sin2h - Cos(2x + h)Sinh

= Sinh{2Cos(4x + 2h)Cosh - Cos(2x + h)}

[Sin2A = 2SinA CosB]

=> $\Delta$(Sinx Cos3x) = Sinh{2Cos(4x + 2h)Cosh - Cos(2x + h)}
 

Question 3: Find the general solution of the equation

$\frac{d^2y}{dx^2}$ - 2$\frac{dy}{dx}$ - 3y = 0
Solution:
Given differential equation

$\frac{d^2y}{dx^2}$ - 2$\frac{dy}{dx}$ - 3y = 0

or D2 y - 2Dy - 3y = 0

or (D2 - 2D - 3)y = 0

or  D2 - 2 D - 3 = 0

Solve for D,

D2 - 2 D - 3 = 0

=> D2 - 3D + D - 3 = 0

=> D(D - 3) + (D - 3) = 0

=> (D + 1)(D - 3) = 0

either D + 1 = 0   or   D - 3 = 0

=> D = -1, 3

The solution of the differential equation is

y = C1 e-x + C2 e3x