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Popular >

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

Solution

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

Solution

y = - \frac{4}{4 t ^{2} - 1}

Solution steps

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

Solve separable ODE:

y = - \frac{4}{4 t ^{2} - 1}

y = - \frac{4}{4 t ^{2} - 1}

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integral of (\sqrt[5]{2x+1})

y^{^{''}} = 9 t^{2}

Frequently Asked Questions (FAQ)

What is the solution for y^'-ty^3=0,y(0)=2 ?
The solution for y^'-ty^3=0,y(0)=2 is y=sqrt(-4/(4t^2-1))