I've just started with differential equations and in the textbook I was given two, with one of which I have trouble. The task was to solve them with software, but I considered it'd be better to solve by hand. But since it's just the start of the topic in the textbook, not much has been provided about how to do it.
The equation and the initial value problem are:
$$ \frac{dy}{dx} = \frac{1}{\sqrt{2^2 - x^2}} \\\ y(0) = 2 $$
I looked into two tables of integrals (so it's doublechecked) and found this:
$$ \int{\frac{1}{\sqrt{a^2 - x^2}}}{dx} = \frac{1}{\sin{\frac{x}{a}}} + C $$
But the problem is that y is not defined at 0 (if the tables are correct). Since y(0) = 2 might be a typo in the textbook, I tried to check whether the solution would work for some other point. I took x = 1 and C = 0 and tried to plot y and t (the tangent) in software:
$$
y(x) = \frac{1}{\sin{\frac{x}{a}}} \\\
y'(1) = \frac{1}{\sqrt{3}} \\\
t(x) = \frac{x - x_0}{\sqrt{3}} + t(x_0) = \frac{x - 1}{\sqrt{3}} + \frac{1}{\sin{\frac{1}{2}}}
$$
As you can see, t (the yellow curve) is definitely not tangent to y(x). But it seems normal, and indeed, when I changed the yellow to $t(x) = -\sqrt{3}(x - 1) + \frac{1}{\sin{\frac{1}{2}}}$ it became tangent at x = 1.
So could anyone, please, explain, what's going on here? Did I use the integral tables correctly? How come the function under the integral sign (which is a derivative) gives not a tangent but normal to the right hand side of the integral solution?
Thank you.
Actually, the basic formula is that $\;\displaystyle\frac1{\sqrt{1-x^2}}$ is the derivative of $\arcsin x$ (or of $-\arccos x$). By the substitution $x=at$, you obtain the standard formula $$\int \frac1{\sqrt{a^2 - x^2}}\,\mathrm dx = \arcsin \frac{x}{a},$$ and you obtain, with the initial condition $y(0)=2$, $$y(x)=2+\int_0^x \frac1{\sqrt{a^2 - t^2}}\,\mathrm dt=2+ \arcsin \frac{x}{a}.$$