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Popular >

y^'=e^{8x+y}

Solution

y^{^{'}} = e^{8 x + y}

Solution

y = - ln (- \frac{1}{8} e^{8 x} - c_{1})

Solution steps

y^{'} (x) = e^{8 x + y}

Solve separable ODE:

y = - ln (- \frac{1}{8} e^{8 x} - c_{1})

y = - ln (- \frac{1}{8} e^{8 x} - c_{1})

Popular Examples

(\partial)/(\partial y)(ze^{xy})

\frac{\partial}{\partial y} (z e^{x y})

limit as x approaches 1 of 3x^3-4x^2+x+2

x \to 1 lim (3 x^{3} - 4 x^{2} + x + 2)

(\partial)/(\partial x)(ln(x^3+y)-z)

\frac{\partial}{\partial x} (ln (x^{3} + y) - z)

integral of (x^{10})/((x^2+x-2))

\int \frac{x ^{10}}{( x ^{2} + x - 2 )} d x

derivative of e^{-1/2}

\frac{d}{d x} (e^{- \frac{1}{2}})

Frequently Asked Questions (FAQ)

What is the solution for y^'=e^{8x+y} ?
The solution for y^'=e^{8x+y} is y=-ln(-1/8 e^{8x}-c_{1})