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Popular >

y^'+y^3=0,y(0)=3

Solution

y^{^{'}} + y^{3} = 0, y (0) = 3

Solution

y = \frac{3}{18 t + 1}

Solution steps

y^{'} (t) + y^{3} = 0

Solve separable ODE:

y = \frac{3}{18 t + 1}

y = \frac{3}{18 t + 1}

Popular Examples

(\partial)/(\partial x)(x^3y^4z^2)

\frac{\partial}{\partial x} (x^{3} y^{4} z^{2})

tangent of y=-(2x)/(x^2+1),(1,-1)tangent of

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

derivative of f(x)=x^3-3x^2+4x-7 derivative of

f (x) = x^{3} - 3 x^{2} + 4 x - 7

d/(dv)(u/(u+v))

\frac{d}{d v} (\frac{u}{u + v})

tangent of-x^3-5x^2+2,\at x=-1 tangent of

- x^{3} - 5 x^{2} + 2, at x = - 1

Frequently Asked Questions (FAQ)

What is the solution for y^'+y^3=0,y(0)=3 ?
The solution for y^'+y^3=0,y(0)=3 is y= 3/(sqrt(18t+1))