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D Y D X = Y Tan X , Y ( 0 ) = 1

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Question

\[\frac{dy}{dx} = y \tan x, y\left( 0 \right) = 1\]
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Solution

\[\frac{dy}{dx} = y \tan x, y\left( 0 \right) = 1\]
\[ \Rightarrow \frac{1}{y}dy = \tan x dx\]
Integrating both sides, we get
\[\int\frac{1}{y}dy = \int\tan x dx\]
\[ \Rightarrow \log \left| y \right| = \log \left| \sec x \right| + C . . . . . (1)\]
We know that at x = 0 and y = 1 . 
Substituting the values of x and y in (1), we get
\[\log \left| 1 \right| = \log \left| 1 \right| + C\]
\[ \Rightarrow C = 0\]
Substituting the value of C in (1), we get
\[\log \left| y \right| = \log \left| \sec x \right| + 0\]
\[ \Rightarrow y = \sec x\]
\[\text{ Hence, }y = \sec x,\text{ where }x \in \left( \frac{- \pi}{2}, \frac{\pi}{2} \right),\text{ is the required solution .}\]

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Chapter 21: Differential Equations - Exercise 22.07 [Page 56]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 21 Differential Equations
Exercise 22.07 | Q 45.1 | Page 56

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