मराठी

Solve the Following Differential Equation: Y E X Y D X = ( X E X Y + Y 2 ) D Y , Y ≠ 0

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प्रश्न

Solve the following differential equation:
\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy, y \neq 0\]

 

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उत्तर

\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy\]
\[ \Rightarrow y e^\frac{x}{y} dx = x e^\frac{x}{y} dy + y^2 dy\]
\[ \Rightarrow y e^\frac{x}{y} dx - x e^\frac{x}{y} dy = y^2 dy\]
\[ \Rightarrow \left( ydx - xdy \right) e^\frac{x}{y} = y^2 dy\]
\[ \Rightarrow \frac{\left( ydx - xdy \right)}{y^2} e^\frac{x}{y} = dy\]
\[ \Rightarrow e^\frac{x}{y} d\left( \frac{x}{y} \right) = dy\]
\[ \Rightarrow \int e^\frac{x}{y} d\left( \frac{x}{y} \right) = \int dy\]
\[ \Rightarrow e^\frac{x}{y} = y + C\]

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पाठ 21: Differential Equations - Exercise 22.07 [पृष्ठ ५५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 21 Differential Equations
Exercise 22.07 | Q 38.3 | पृष्ठ ५५

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