Question

Verify that $\tt y=\dfrac{a}{x}+b$ is a solution of the differential equation
$\boxed{ \red{ \pmb{ \tt \dfrac{d^{2} y}{d x^{2}}+\dfrac{2}{x}\left(\dfrac{d y}{d x}\right)=0 }}}$
​

Answer 1

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

Given curve is

$\rm \: y=\dfrac{a}{x}+b$

can be rewritten as

$\rm \: y = \dfrac{a + bx}{x}$

$\rm \: xy = bx + a$

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

$\rm \: \dfrac{d}{dx}xy = \dfrac{d}{dx}(bx + a)$

We know,

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx}uv \: = \: v\dfrac{d}{dx}u \: + \: u\dfrac{d}{dx}v \: \: }}$

and

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx}x \: = \: 1 \: }} \: \: \rm \: \: and \: \: \boxed{\sf{  \:\rm \: \dfrac{d}{dx}k = 0 \: \: }}$

So, using these results, we get

$\rm \: x\dfrac{dy}{dx} + y\dfrac{d}{dx}x = b \times 1 + 0$

$\rm \: x\dfrac{dy}{dx} + y = b$

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

$\rm \:\dfrac{d}{dx}\bigg( x\dfrac{dy}{dx} + y\bigg) =\dfrac{d}{dx} b$

$\rm \: x\dfrac{d}{dx}\bigg(\dfrac{dy}{dx}\bigg) + \dfrac{dy}{dx} \times \dfrac{d}{dx}x + \dfrac{dy}{dx} = 0$

$\rm \: x\dfrac{ {d}^{2}y }{d {x}^{2} } + \dfrac{dy}{dx} \times 1 + \dfrac{dy}{dx} = 0$

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

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

On dividing both sides by x, we get

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

Hence,

$\rm\implies \:\boxed{ { \pmb{ \tt \dfrac{d^{2} y}{d x^{2}}+\dfrac{2}{x}\left(\dfrac{d y}{d x}\right)=0 }}}$

[ See the alternative solution attached in attachment ]

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Additional Information :-

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