Question

यदि y=500e^{7x} +600e^{-7x} है तो दर्शाइए कि तट d^{2}y/dx^{2} = 49y है।

Answer 1

$\huge \mathfrak{solution}$

Given:

$y = 500 {e}^{7x} + 600 {e}^{ - 7x}$

To prove :

$\frac{ {d}^{2} y}{d {x}^{2} } = 49y$

$\implies \: y = 500 {e}^{7x} + 600 {e}^{ - 7x}$

Differentiate w.r.t. x

$\implies \: \frac{dy}{dx} = 500 \frac{d}{dx} {e}^{7x} + 600 \frac{d}{dx} {e}^{ - 7x} \\ \\ \implies \: \frac{dy}{dx} = 500 \times 7 {e}^{7x} + 600 \times - 7 {e}^{ - 7x} \\ \\ \implies \frac{dy}{dx} =( 7)500 {e}^{7x} + ( -7)600 {e}^{ - 7x}$

Again differentiate w.r.t. x ,

$\star \: \frac{ {d}^{2}y }{d {x}^{2} } = \frac{d}{dx} ((7 )\times 500 {e}^{7x} +( - 7)600 {e}^{ - 7x} ) \\ \\ \star \frac{ {d}^{2} y}{d {x}^{2} } = (7)(7)500 {e}^{7x} + ( - 7)( - 7)600 {e}^{ - 7x} \\ \\ \star \frac{ {d}^{2} y}{d {x}^{2} } = (49)500 {e}^{7x} + (49)600 {e}^{ - 7x} \\ \\ \star \: \frac{ {d}^{2}y }{d {x}^{2} } = 49(500 {e}^{7x} + 600 {e}^{ - 7x} )$

And

y = 500e^7x +600e^-7x

$\star \: \frac{ {d}^{2} y}{dx ^{2} } = 49y$

proved

Formula used:

$\huge \star \bold{ \frac{d}{dx} {e}^{ax} = {e }^{ax} (a)}$

Answer 2

d²y/dx² = 49y यदि y = 500e⁷ˣ  + 600e⁻⁷ˣ

Step-by-step explanation:

y = 500e⁷ˣ  + 600e⁻⁷ˣ

dy/dx = 500e⁷ˣ * 7  + 600e⁻⁷ˣ *(-7)

=> dy/dx = 7 ( 500e⁷ˣ - 600e⁻⁷ˣ)

=> d²y/dx² = 7 (  500e⁷ˣ * 7  - 600e⁻⁷ˣ *(-7))

=> d²y/dx² = 7 (  500e⁷ˣ * 7  + 600e⁻⁷ˣ *(7))

=>  d²y/dx² = 7 * 7(  500e⁷ˣ  + 600e⁻⁷ˣ )

=>  d²y/dx² = 49(  500e⁷ˣ  + 600e⁻⁷ˣ )

=> d²y/dx² = 49y

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