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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

y^{^{'}} = - y \cdot cos (x)

y^2sec(5x)(dy)/(dx)=1,y(0)=0

y^{2} sec (5 x) \frac{d y}{d x} = 1, y (0) = 0

\frac{d y}{d t} = 8 - y

(dy)/(dx)=(-e^x(y-1))/(2(e^x+4))

\frac{d y}{d x} = \frac{- e ^{x} ( y - 1 )}{2 ( e ^{x} + 4 )}

y^'cos(y)+2xsin(y)=-2x

y^{^{'}} cos (y) + 2 x sin (y) = - 2 x

Frequently Asked Questions (FAQ)

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