मराठी

Solve the Following Initial Value Problem: D Y D X − Y X + C O S E C Y X = 0 , Y ( 1 ) = 0

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प्रश्न

Solve the following initial value problem:
\[\frac{dy}{dx} - \frac{y}{x} + cosec\frac{y}{x} = 0, y\left( 1 \right) = 0\]

बेरीज
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उत्तर

\[\frac{dy}{dx} - \frac{y}{x} + cosec\frac{y}{x} = 0, y\left( 1 \right) = 0\]
This is an homogenous equation, put y = vx
\[\frac{dy}{dx} + v + x\frac{dv}{dx}\]
\[v + x\frac{dv}{dx} - v + cosec\ v = 0\]
\[x\frac{dv}{dx} = cosec\ v\]
\[\frac{dv}{cosec\ v} = \frac{dx}{x}\]
\[\sin v\ dv = \frac{dx}{x}\]
On integrating both sides, we get
\[\int \sin v\ dv = \int\frac{dx}{x}\]
\[ - \cos v = \log_e x + c\]
\[ - \cos v + \log_e x = c\]
\[\cos v + \log_e x = - c\]
\[\cos \left( \frac{y}{x} \right) + \log_e x = - c\]
\[\text{ As }y\left( 1 \right) = 0\]
\[\cos \left( \frac{0}{1} \right) = 0 + \log_e 1 = - c\]
\[1 + 0 = - c\]
\[ \Rightarrow c = - 1\]
\[ \Rightarrow \cos \left( \frac{y}{x} \right) + \log_e x = 1\]
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पाठ 21: Differential Equations - Exercise 22.09 [पृष्ठ ८४]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 21 Differential Equations
Exercise 22.09 | Q 36.3 | पृष्ठ ८४

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