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Popular >

y^'-e^{2x+y}=0

Solution

y^{^{'}} - e^{2 x + y} = 0

Solution

y = - ln (- \frac{1}{2} e^{2 x} - c_{1})

Solution steps

y^{'} (x) - e^{2 x + y} = 0

Solve separable ODE:

y = - ln (- \frac{1}{2} e^{2 x} - c_{1})

y = - ln (- \frac{1}{2} e^{2 x} - c_{1})

Popular Examples

integral of 1-cos(6x)

\int 1 - cos (6 x) d x

limit as t approaches 1 of (1/t-1)/(t-1)

t \to 1 lim (\frac{\frac{1}{t} - 1}{t - 1})

sum from n=0 to infinity of n^2

n = 0 \sum \infty n^{2}

derivative of ln(x^2-y)

\frac{d}{d x} (ln (x^{2} - y))

derivative of f(x)=(4x+5)(5x-6)derivative of

f (x) = (4 x + 5) (5 x - 6)

Frequently Asked Questions (FAQ)

What is the solution for y^'-e^{2x+y}=0 ?
The solution for y^'-e^{2x+y}=0 is y=-ln(-1/2 e^{2x}-c_{1})