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Popular >

y^'-2ty^3=0,y(2)=1

Solution

y^{^{'}} - 2 t y^{3} = 0, y (2) = 1

Solution

y = - \frac{1}{2 t ^{2} - 9}

Solution steps

y^{'} (t) - 2 t y^{3} = 0

Solve separable ODE:

y = - \frac{1}{2 t ^{2} - 9}

y = - \frac{1}{2 t ^{2} - 9}

Popular Examples

y^{^{'}} = \frac{7 y}{x}

(dy)/(dx)=(xsqrt(1-y^2))/(ysqrt(2x^2+1))

\frac{d y}{d x} = \frac{x 1 - y ^{2}}{y 2 x ^{2} + 1}

(dy)/(dx)=y(2+sin(x))

\frac{d y}{d x} = y (2 + sin (x))

y^'=(x^2+1)/(2-y)

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

y^'=(2x+xy^2)/(4y+yx^2)

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

Frequently Asked Questions (FAQ)

What is the solution for y^'-2ty^3=0,y(2)=1 ?
The solution for y^'-2ty^3=0,y(2)=1 is y=sqrt(-1/(2t^2-9))