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Popular >

y^'=te^{-y},y(0)=1

Solution

y^{^{'}} = t e^{- y}, y (0) = 1

Solution

y = ln (\frac{t ^{2}}{2} + e)

Solution steps

y^{'} (t) = t e^{- y}

Solve separable ODE:

y = ln (\frac{t ^{2}}{2} + e)

y = ln (\frac{t ^{2}}{2} + e)

Popular Examples

integral from 0 to 1 of cos(2pix)

\int_{0}^{1} cos (2 π x) d x

limit as x approaches 3 of x-1

x \to 3 lim (x - 1)

y^{''}+36y=-48sin(6t)

y^{^{''}} + 36 y = - 48 sin (6 t)

derivative of x^2+x-6

\frac{d}{d x} (x^{2} + x - 6)

derivative of f(x)=(3x^2-4x)e^x derivative of

f (x) = (3 x^{2} - 4 x) e^{x}

Frequently Asked Questions (FAQ)

What is the solution for y^'=te^{-y},y(0)=1 ?
The solution for y^'=te^{-y},y(0)=1 is y=ln((t^2)/2+e)