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Let the equation of the plane be

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

where a^{2} + b^{2} + c^{2}= k^{2 } ......................(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 = k^{2/3} x^{1/3}, b = k^{2/3} y^{1/3}, c = k^{2/3} z^{1/3}

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

x^{2/3} + y^{2/3} + z^{2/3} = k^{2/3 }= Constant

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

where a

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 = k

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

x

$\Delta$(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)}

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

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

Given differential equation

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

or D^{2} y - 2Dy - 3y = 0

or (D^{2} - 2D - 3)y = 0

or D^{2} - 2 D - 3 = 0

Solve for D,

D^{2} - 2 D - 3 = 0

=> D^{2} - 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 = C**_{1} e^{-x} + C_{2} e^{3x}

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

or D

or (D

or D

Solve for D,

D

=> D

=> 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