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Popular >

y^'=1-t+y^2-ty^2

Solution

y^{^{'}} = 1 - t + y^{2} - t y^{2}

Solution

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

Solution steps

y^{'} (t) = 1 - t + y^{2} - t y^{2}

Solve separable ODE:

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

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

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derivative of cos(2x*sin(2x))

\frac{d}{d x} (cos (2 x) \cdot sin (2 x))

integral of 6/(x^5)

\int \frac{6}{x ^{5}} d x

Frequently Asked Questions (FAQ)

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