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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

(dy)/(dx)=cos^2(+sin(2x))

\frac{d y}{d x} = cos^{2} (+ sin (2 x))

integral of (x+7)e^{2x+3}

\int (x + 7) e^{2 x + 3} d x

integral from-2 to 3 of (33)/(x^4)

\int_{- 2}^{3} \frac{33}{x ^{4}} d x

integral of x^3-6x^2+x+3

\int x^{3} - 6 x^{2} + x + 3 d x

limit as x approaches-7 of (x^2-5)/(7-x)

x \to - 7 lim (\frac{x ^{2} - 5}{7 - 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})