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Popular >

y^'=(sqrt(1-y^2))/t

Solution

y^{^{'}} = \frac{1 - y ^{2}}{t}

Solution

y = sin (ln (t) + c_{1})

Solution steps

y^{'} (t) = \frac{1 - y ^{2}}{t}

Solve separable ODE:

y = sin (ln (t) + c_{1})

y = sin (ln (t) + c_{1})

Popular Examples

tangent of f(x)=4x^2+2x,(-2,12)tangent of

f (x) = 4 x^{2} + 2 x, (- 2, 12)

integral of (x^7)/2-1/(x^5)

\int \frac{x ^{7}}{2} - \frac{1}{x ^{5}} d x

limit as n approaches infinity of sin(2n)

n \to \infty lim (sin (2 n))

taylor e^{-3x},0 taylor

e^{- 3 x}, 0

sum from n=2 to infinity of x^n

n = 2 \sum \infty x^{n}

Frequently Asked Questions (FAQ)

What is the solution for y^'=(sqrt(1-y^2))/t ?
The solution for y^'=(sqrt(1-y^2))/t is y=sin(ln(t)+c_{1})