Question

y=ax²+b formation of differential equations by eliminating orbitary constants​

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm \: y = {ax}^{2} + b - - - (1)$

Since, the given curve has two arbitrary constants a and b, so we have to differentiate twice the given equation and then we have to eliminate a and b to get the required differential equation.

So, on differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}y = \dfrac{d}{dx}( {ax}^{2} + b)$

$\rm \: \dfrac{dy}{dx} = a\dfrac{d}{dx} {x}^{2} + \dfrac{d}{dx}b$

$\rm \: \dfrac{dy}{dx} = 2ax + 0$

$\rm \: \dfrac{dy}{dx} = 2ax - - - - (2)$

On differentiating both sides w. r. t. x, we get

$\rm \:\dfrac{d}{dx}\bigg( \dfrac{dy}{dx}\bigg) = \dfrac{d}{dx}2ax$

$\rm \: \dfrac{ {d}^{2} y}{ {dx}^{2} } = 2a\dfrac{d}{dx}x$

$\rm \: \dfrac{ {d}^{2} y}{ {dx}^{2} } = 2a \times 1$

$\rm \: \dfrac{ {d}^{2} y}{ {dx}^{2} } = 2a - - - (3)$

On substituting the value of 2a in equation (2), we get

$\rm \:\dfrac{dy}{dx} = x\dfrac{ {d}^{2} y}{ {dx}^{2} }$

$\rm\implies \:\rm \: x\dfrac{ {d}^{2} y}{ {dx}^{2} } - \dfrac{dy}{dx} = 0$

is the required differential equation.

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Formulae used :-

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: \: }}$

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx} k = 0 \: \: }}$

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx}x = 1 \: \: }}$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$