MHB Solve the initial value problem

[karush]

Solve the initial value problem
for $y$ as a function of $x$
\begin{align*}\displaystyle
\sqrt{16-x^2} \, \frac{dy}{dx}&=1, \, x<4, y(0)=12
\end{align*}

assume the first thing to do is $\int$ both sides

 

[MarkFL]

I would separate variables, switch dummy variables, and use the boundaries:

$$\int_{y_0}^{y(x)}\,du=\int_0^x \left(16-v^2\right)^{-\frac{1}{2}}\,dv$$

 

[Prove It]

Solve the initial value problem
for $y$ as a function of $x$
\begin{align*}\displaystyle
\sqrt{16-x^2} \, \frac{dy}{dx}&=1, \, x<4, y(0)=12
\end{align*}

assume the first thing to do is $\int$ both sides

I don't see why you would need dummy variables...

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{\sqrt{16 - x^2}} \\ y &= \int{ \frac{1}{\sqrt{16 - x^2}}\,\mathrm{d}x } \end{align*}$

Once you integrate you will have a constant of integration, use your initial condition to determine it.

 

[HallsofIvy]

You don't need them but it helps to distinguish the integrand from the final integral.

 

[karush]

the MML sample didn't use dummy vars
but showing 20+ steps I got lost. so want see how this goes
$$\int_{y_0}^{y(x)}\,du=\int_0^x \left(16-v^2\right)^{-\frac{1}{2}}\,dv$$
so next
$\displaystyle
\left[u\right]_{12}^{y(x)}
=\left[\arcsin\left(\frac{u}{4}\right)\right]_0^x$
kinda ??