Question

Solve the initial - value problem.

Answer 1

\[\frac{ dy }{ dx }=\frac{ xtany }{ 1+\sqrt{x} }, y(0)=1\]

Answer 2

\[
\frac {dy}{\tan(y)} = \frac { x dx}{1+\sqrt x }
\]
Can you finish it?

Answer 3

Of coarse

Answer 4

\[
\int \frac {dy}{\tan(y)} = \int\frac { x dx}{1+\sqrt x }\\
\ln (\sin (y))=\frac{2 x^{3/2}}{3}-x+2 \sqrt{x}-2 \log \left(\sqrt{x}+1\right)+C\\
\ln(\sin(1))=C


\]

Answer 5

Thanks :)

Answer 6

YW