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Popular >

y^'=1-t+y-ty

Solution

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

Solution

y = e^{- \frac{t ^{2}}{2} + t + c_{1}} - 1

Solution steps

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

Solve separable ODE:

y = e^{- \frac{t ^{2}}{2} + t + c_{1}} - 1

y = e^{- \frac{t ^{2}}{2} + t + c_{1}} - 1

Popular Examples

integral of (ln^3(x^2))/x

\int \frac{ln ^{3} ( x ^{2} )}{x} d x

derivative of (6x)/(4+x^2)derivative of

\frac{6 x}{4 + x ^{2}}

(\partial)/(\partial x)(cos(6x))

\frac{\partial}{\partial x} (cos (6 x))

derivative of |x^3+8x^2+5x-50|

\frac{d}{d x} (x^{3} + 8 x^{2} + 5 x - 50)

integral of 1/(2+8x)

\int \frac{1}{2 + 8 x} d x

Frequently Asked Questions (FAQ)

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