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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Graph

Popular Examples

y^'-x^5y^4=0,y(0)=2,y(2)=a

y^{^{'}} - x^{5} y^{4} = 0, y (0) = 2, y (2) = a

2y^'-yy^'=x^2+1

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

(2dy)/(dt)+4y=6,y(0)=1(1/2)

\frac{2 d y}{d t} + 4 y = 6, y (0) = 1 (\frac{1}{2})

(dy)/(dx)=(2-y)sin(x),x=2,c^1=3

\frac{d y}{d x} = (2 - y) sin (x), x = 2, c^{1} = 3

x^'+2^2*3^2*x=0

x^{^{'}} + 2^{2} \cdot 3^{2} \cdot x = 0

Frequently Asked Questions (FAQ)

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