MHB Solving for dy/dx: $\sqrt{y}\cos^2{\sqrt{y}}$

[karush]

$\displaystyle
\d{y}{x}=\frac{1}{7}\sqrt{y}\cos^2{\sqrt{y}} \\
$
then
$\displaystyle dy=\frac{1}{7}\sqrt{y}\cos^2{\sqrt{y}} \, dx$
and then
$\displaystyle \frac{dy}{\frac{1}{7}\sqrt{y}\cos^2{\sqrt{y}} }=\frac{1}{7} \, dx$
and then ?

 

[Prove It]

$\displaystyle
\d{y}{x}=\frac{1}{7}\sqrt{y}\cos^2{\sqrt{y}} \\
$
then
$\displaystyle dy=\frac{1}{7}\sqrt{y}\cos^2{\sqrt{y}} \, dx$
and then
$\displaystyle \frac{dy}{\frac{1}{7}\sqrt{y}\cos^2{\sqrt{y}} }=\frac{1}{7} \, dx$
and then ?

First of all, as you have kept the $\displaystyle \begin{align*} \frac{1}{7} \end{align*}$ on the RHS you should NOT have divided it on the left. You should have

$\displaystyle \begin{align*} \int{ \frac{\mathrm{d}y}{\sqrt{y}\,\cos^2{\left( \sqrt{y} \right) }} } &= \int{ \frac{1}{7}\,\mathrm{d}x } \\ \int{ \frac{\mathrm{d}y}{2\,\sqrt{y}\,\cos^2{ \left( \sqrt{y} \right) } } } &= \int{ \frac{1}{14}\,\mathrm{d}x } \end{align*}$

Substitute $\displaystyle \begin{align*} u = \sqrt{y} \implies \mathrm{d}u = \frac{\mathrm{d}y}{2\,\sqrt{y}} \end{align*}$ giving

$\displaystyle \begin{align*} \int{ \frac{\mathrm{d}u}{\cos^2{ \left( u \right) } } } &= \int{ \frac{1}{14}\,\mathrm{d}x } \end{align*}$

These should both be easy to integrate now.

 

[karush]

\begin{align*}\displaystyle
\int{ \frac{{d}u}{\cos^2{ \left( u \right) } } } &= \int{ \frac{1}{14}\, {d}x }
\end{align*}
$\textit{using the standard integrals}$

$\displaystyle\tan\left({u}\right)=\frac{x}{14}$

$u = \sqrt{y}$

$\displaystyle\tan\left({ \sqrt{y}}\right)=\frac{x}{14}$

$\textit{done ?}$

 

[Prove It]

\begin{align*}\displaystyle
\int{ \frac{{d}u}{\cos^2{ \left( u \right) } } } &= \int{ \frac{1}{14}\, {d}x }
\end{align*}
$\textit{using the standard integrals}$

$\displaystyle\tan\left({u}\right)=\frac{x}{14}$

$u = \sqrt{y}$

$\displaystyle\tan\left({ \sqrt{y}}\right)=\frac{x}{14}$

$\textit{done ?}$

Plus a constant.

 

[karush]

Plus a constant.

$\displaystyle\tan\left({ \sqrt{y}}\right)=\frac{x}{14}+C$

really?

 

[Prove It]

$\displaystyle\tan\left({ \sqrt{y}}\right)=\frac{x}{14}+C$

really?

Technically as you have integrated on both sides, both sides need an integration constant, but since they can then be combined on one side, yes what you have here is correct.

You could then solve for y if you really wanted to...

 

[karush]

too lazy.. I leave it the way it is.