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Popular >

y^'x=sqrt(1-y^2)

Solution

y^{^{'}} x = 1 - y^{2}

Solution

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

Solution steps

y^{'} (x) x = 1 - y^{2}

Solve separable ODE:

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

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

Graph

Popular Examples

y^'=(2000-2y)/(200-x),y(0)=0

y^{^{'}} = \frac{2000 - 2 y}{200 - x}, y (0) = 0

16y-8y^'+145y=0,y(0)=1

16 y - 8 y^{^{'}} + 145 y = 0, y (0) = 1

e^yy^'=((ln(t))^2)/t ,y(1)=ln(7)

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}, y (1) = ln (7)

f^'(x)=0.039f(x)

f^{^{'}} (x) = 0.039 f (x)

y^{^{'}} = \frac{- x y ^{2}}{4}

Frequently Asked Questions (FAQ)

What is the solution for y^'x=sqrt(1-y^2) ?
The solution for y^'x=sqrt(1-y^2) is y=sin(ln(x)+c_{1})