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Popular >

y^'=e^{t+y}

Solution

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

Solution

y = - ln (- e^{t} - c_{1})

Solution steps

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

Solve separable ODE:

y = - ln (- e^{t} - c_{1})

y = - ln (- e^{t} - c_{1})

Popular Examples

(\partial)/(\partial x)(-3cos(x))

\frac{\partial}{\partial x} (- 3 cos (x))

inverse oflaplace 1/(s(12s^2+10s+1))inverse laplace

\frac{1}{s ( 12 s ^{2} + 10 s + 1 )}

integral from 0 to 4 of (4x)^2-(x^2)^2

\int_{0}^{4} (4 x)^{2} - (x^{2})^{2} d x

(\partial)/(\partial y)(ln(sqrt(3x)+5y))

\frac{\partial}{\partial y} (ln (3 x + 5 y))

derivative of 5e^{-5x}

\frac{d}{d x} (5 e^{- 5 x})

Frequently Asked Questions (FAQ)

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