मराठी

Solve the Following Initial Value Problem: (Xy − Y2) Dx − X2 Dy = 0, Y(1) = 1

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प्रश्न

Solve the following initial value problem:
(xy − y2) dx − x2 dy = 0, y(1) = 1

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उत्तर

(xy − y2) dx − x2 dy = 0, y(1) = 1
This is an homogenous equation, put y = vx
\[\frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[\left( xy - y^2 \right) = x^2 \left( \frac{dy}{dx} \right)\]
\[\left( v x^2 - v^2 x^2 \right) = x^2 \left( v + x\frac{dv}{dx} \right)\]
\[v x^2 \left( 1 - v \right) = x^2 \left( v + x\frac{dv}{dx} \right)\]
\[v\left( 1 - v \right) = v + x\frac{dv}{dx}\]
\[v - v^2 = v + x\frac{dv}{dx}\]
\[ - v^2 = x\frac{dv}{dx}\]
\[ - \frac{1}{x}dx = \frac{1}{v^2}dv\]
On integrating both sides we get,
\[- \int\frac{1}{x}dx = \int\frac{1}{v^2}dv\]
\[ - \log_e x = \frac{v^{- 2 + 1}}{- 2 + 1} + c\]
\[ - \log_e x = \frac{v^{- 1}}{- 1} + c\]
\[ - \log_e x = - \frac{1}{v} + c\]
\[ - \log_e x = - \frac{1}{v} + c\]
\[\frac{x}{y} - \log_e x = c\]
\[\text{ As }y\left( 1 \right) = 1\]
\[\frac{1}{1} - \log_e 1 = c\]
\[ \Rightarrow c = 1\]

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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 36.4 | पृष्ठ ८४

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