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

• MHB
• karush
In summary, the conversation discusses finding the solution of initial value and interval for the given equation, which involves separating and integrating. The method of trig substitution is assumed to be used to solve the integral. The book provides the detailed steps to solve the equation, and the conversation reflects on the tediousness of the process. Finally, the final solution is given as y = [(3(arcsin(x))^2+2)/2]^(1/3).
karush
Gold Member
MHB
(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

Last edited:
karush said:
(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:

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

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

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

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

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

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

sure like the boundary method

most examples just plow thru another 5 tedious steps

karush said:
(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 ...

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

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

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

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

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

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

## 1. What does the "-2.2.20" in the IVP interval mean?

The "-2.2.20" in the IVP interval refers to the date when the initial value problem (IVP) was created or solved. It is common practice in scientific and mathematical notation to include a date or version number as a way to track and reference different versions of a problem.

## 2. What is an IVP interval?

An IVP interval is a specific range of values for the independent variable in an initial value problem. In this case, the IVP interval is from -2 to 2, indicating that the problem is being solved for values of x between -2 and 2.

## 3. What does "trig subst" mean in this equation?

"Trig subst" is short for "trigonometric substitution", which is a technique used in calculus to simplify integrals involving trigonometric functions. In this problem, the "trig subst" indicates that a trigonometric substitution will be used to solve the integral.

## 4. What is the meaning of the symbols and numbers in this equation?

The symbols and numbers in this equation have specific mathematical meanings. The "y" and "x" represent the dependent and independent variables, respectively. The "^" symbol indicates exponentiation, and the "1/2" represents taking the square root. The "dy" and "dx" indicate that the integral is being solved with respect to y and x, respectively. The "\arcsin{x}" represents the inverse sine function, and the "(1-x^2)^{1/2}" represents the Pythagorean identity for sine.

## 5. How do you solve this initial value problem using trigonometric substitution?

To solve this initial value problem, you would first substitute the given equation for y into the integral, using the trigonometric substitution y = sinθ. Then, you would use trigonometric identities to simplify the integral and solve for θ. Finally, you would use inverse trigonometric functions to solve for x and find the solution to the IVP.

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