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Popular >

y^'=t+ty^2

Solution

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

Solution

y = tan (\frac{t ^{2}}{2} + c_{1})

Solution steps

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

Solve separable ODE:

y = tan (\frac{t ^{2}}{2} + c_{1})

y = tan (\frac{t ^{2}}{2} + c_{1})

Popular Examples

tangent of f(x)=x+sqrt(x),\at x=4 tangent of

f (x) = x + x, at x = 4

integral from-1 to 4 of (x^2+2x-1)

\int_{- 1}^{4} (x^{2} + 2 x - 1) d x

derivative of-yx

\frac{d}{d x} (- y x)

49y^{''}+140y^'+100y=0

49 y^{^{''}} + 140 y^{^{'}} + 100 y = 0

derivative of sqrt(x^2+3)derivative of

x^{2} + 3

Frequently Asked Questions (FAQ)

What is the solution for y^'=t+ty^2 ?
The solution for y^'=t+ty^2 is y=tan((t^2)/2+c_{1})