-2.2.20 IVP interval....trig subst y^2(1-x^2)^{1/2} \,dy=\arcsin{x}\,dx

[karush]

(a) find solution of initial value and (c) interval
$$\quad\displaystyle
y^2(1-x^2)^{1/2} \,dy = \arcsin{x}\,dx,
\quad y(0) = 1$$
separate
$$y^2 \,dy = \frac{\arcsin{x}}{(1-x^2)^{1/2}}\,dx,$$
Integrate
\begin{align*}
\int y^2 \,dy& = \int\frac{\arcsin{x}}{\sqrt{(1-x^2)}}\,dx,
\end{align*}
ok I assume a trig substitution to solvebook answer

 

[MarkFL]

(a) find solution of initial value and (c) interval
$$\quad\displaystyle
y^2(1-x^2)^{1/2} \,dy = \arcsin{x}\,dx,
\quad y(0) = 1$$
separate
$$y^2 \,dy = \frac{\arcsin{x}}{(1-x^2)^{1/2}}\,dx,$$
Integrate
\begin{align*}
\int y^2 \,dy& = \int\frac{\arcsin{x}}{\sqrt{(1-x^2)}}\,dx,
\end{align*}
ok I assume a trig substitution to solve

Yes, I would let:

$$\theta=\arcsin(x)\implies x=\sin(\theta)\implies dx=\cos(\theta)\,d\theta$$

And then we have after simplification, and using the given boundaries:

$$\int_1^y u^2\,du=\int_0^{\arcsin(x)} v\,dv$$

$$\frac{1}{3}(y^3-1)=\frac{1}{2}\arcsin^2(x)$$

$$y^3=\frac{3}{2}\arcsin^2(x)+1$$

$$y=\left(\frac{3}{2}\arcsin^2(x)+1\right)^{\Large\frac{1}{3}}$$

 

[karush]

sure like the boundary method

most examples just plow thru another 5 tedious steps

 

[skeeter]

(a) find solution of initial value and (c) interval
$$\quad\displaystyle
y^2(1-x^2)^{1/2} \,dy = \arcsin{x}\,dx,
\quad y(0) = 1$$
separate
$$y^2 \,dy = \frac{\arcsin{x}}{(1-x^2)^{1/2}}\,dx,$$
Integrate
\begin{align*}
\int y^2 \,dy& = \int\frac{\arcsin{x}}{\sqrt{(1-x^2)}}\,dx,
\end{align*}
ok I assume a trig substitution to solvebook answer

$\displaystyle \int y^2 \, dy = \int \dfrac{\arcsin{x}}{\sqrt{(1-x^2)}} \, dx$

RHS ...

$u = \arcsin{x}$, $du = \dfrac{dx}{\sqrt{1-x^2}}$

$\displaystyle \int y^2 \, dy = \int u \, du$

$\dfrac{y^3}{3} = \dfrac{u^2}{2} + C$

$y(0) = 1 \implies u = 0 \implies C = \dfrac{1}{3}$ ...

$\dfrac{y^3}{3} = \dfrac{3u^2+2}{6}$

$y = \left[\dfrac{3(\arcsin{x})^2+2}{2}\right]^{1/3}$

 

[karush]

I always have struggles with these trig subst...

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