આપેલ છે કે frac{d^2 y}{d x^2}+cot x frac{d y} | vedclass

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(D) આદેશ $z = \log 	an \frac{x}{2}$ આપેલ છે.તેથી, $\frac{dz}{dx} = \frac{1}{	an(x/2)} \cdot \sec^2(x/2) \cdot \frac{1}{2} = \frac{1}{\sin x} = \oper

Vedclass · Accepted Answer

$\frac{d^2 y}{d z^2}+4 y=0$

Vedclass · Answer

(D) આદેશ $z = \log 	an \frac{x}{2}$ આપેલ છે.તેથી, $\frac{dz}{dx} = \frac{1}{	an(x/2)} \cdot \sec^2(x/2) \cdot \frac{1}{2} = \frac{1}{\sin x} = \operatorname{cosec} x$.હવે, $\frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx} = \operatorname{cosec} x \frac{dy}{dz}$.આગળ, $\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \operatorname{cosec} x \frac{dy}{dz} ight) = \frac{d}{dz} \left( \operatorname{cosec} x \frac{dy}{dz} ight) \cdot \frac{dz}{dx} = \left( -\operatorname{cosec} x \cot x \frac{dy}{dz} + \operatorname{cosec} x \frac{d^2 y}{dz^2} ight) \operatorname{cosec} x = \operatorname{cosec}^2 x \frac{d^2 y}{dz^2} - \operatorname{cosec} x \cot x \frac{dy}{dz}$.આ કિંમતોને મૂળ સમીકરણ $\frac{d^2 y}{dx^2} + \cot x \frac{dy}{dx} + 4y \operatorname{cosec}^2 x = 0$ માં મૂકતા:$\left( \operatorname{cosec}^2 x \frac{d^2 y}{dz^2} - \operatorname{cosec} x \cot x \frac{dy}{dz} ight) + \cot x (\operatorname{cosec} x \frac{dy}{dz}) + 4y \operatorname{cosec}^2 x = 0$.$\operatorname{cosec}^2 x \frac{d^2 y}{dz^2} + 4y \operatorname{cosec}^2 x = 0$.$\operatorname{cosec}^2 x$ વડે ભાગતા, આપણને $\frac{d^2 y}{dz^2} + 4y = 0$ મળે છે.

Similar Questions

વિકલ સમીકરણ $x \frac{d^2 y}{d x^2} = 1$ નો ઉકેલ,જ્યાં $x = y = 1$ અને $x = 1$ આગળ $\frac{dy}{d x} = 0$ હોય,તે શોધો:

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