Question

\[\sin\left( \frac{dy}{dx} \right) = k ; y\left( 0 \right) = 1\]

Answer 1

We have, 
\[\sin \frac{dy}{dx} = k\]
\[ \Rightarrow \frac{dy}{dx} = \sin^{- 1} k\]
\[ \Rightarrow dy = \left\{ \sin^{- 1} k \right\}dx\]
Integrating both sides, we get
\[\int dy = \int\left( \sin^{- 1} k \right) dx\]
\[ \Rightarrow y = x \sin^{- 1} k + C . . . . . \left( 1 \right)\]
\[ \text{ It is given that }y\left( 0 \right) = 1 . \]
\[ \therefore 1 = 0 \times \sin^{- 1} k + C\]
\[ \Rightarrow C = 1\]
\[\text{ Substituting the value of C in }\left( 1 \right),\text{ we get }\]
\[y = x \sin^{- 1} k + 1\]
\[ \Rightarrow y - 1 = x \sin^{- 1} k \]
\[\text{ Hence, }y - 1 = x \sin^{- 1} \text{ k is the solution to the given differential equation.}\]