- #1

karush

Gold Member

MHB

- 3,269

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(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

$$\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

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