मराठी

√ a + X D Y + X D X = 0

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प्रश्न

\[\sqrt{a + x} dy + x\ dx = 0\]
बेरीज
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उत्तर

We have, 
\[\sqrt{a + x}dy + x\ dx = 0\]
\[ \Rightarrow \sqrt{a + x}dy = - xdx\]
\[ \Rightarrow dy = \frac{- x}{\sqrt{a + x}}dx\]
\[ \Rightarrow dy = - \frac{\left( x + a - a \right)}{\sqrt{a + x}}dx\]
\[ \Rightarrow dy = - \left( \sqrt{a + x} - \frac{a}{\sqrt{a + x}} \right)dx\]
Integrating both sides, we get
\[\int dy = - \int\left( \sqrt{a + x} - \frac{a}{\sqrt{a + x}} \right)dx\]
\[ \Rightarrow y = - \frac{2 \left( a + x \right)^\frac{3}{2}}{3} + 2a\sqrt{a + x} + C\]
\[ \Rightarrow y + \frac{2}{3} \left( a + x \right)^\frac{3}{2} - 2a\sqrt{a + x} = C\]
\[\text{ Hence, }y + \frac{2}{3} \left( a + x \right)^\frac{3}{2} - 2a\sqrt{a + x} = \text{C is the solution to the given differential equation.}\]

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पाठ 21: Differential Equations - Exercise 22.05 [पृष्ठ ३४]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 21 Differential Equations
Exercise 22.05 | Q 17 | पृष्ठ ३४

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

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