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Popular >

y^'=t+ty^2,y(2)=0

Solution

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

Solution

y = tan (\frac{t ^{2}}{2} - 2)

Solution steps

y^{'} (t) = t + t y^{2}

Solve separable ODE:

y = tan (\frac{t ^{2}}{2} - 2)

y = tan (\frac{t ^{2}}{2} - 2)

Graph

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\int y^{4} - 8 y^{3} + 16 y^{2} d y

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Frequently Asked Questions (FAQ)

What is the solution for y^'=t+ty^2,y(2)=0 ?
The solution for y^'=t+ty^2,y(2)=0 is y=tan((t^2)/2-2)