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Popular >

y^'-e^{9x+y}=0

Solution

y^{^{'}} - e^{9 x + y} = 0

Solution

y = - ln (- \frac{1}{9} e^{9 x} - c_{1})

Solution steps

y^{'} (x) - e^{9 x + y} = 0

Solve separable ODE:

y = - ln (- \frac{1}{9} e^{9 x} - c_{1})

y = - ln (- \frac{1}{9} e^{9 x} - c_{1})

Popular Examples

integral of 2cos^4(3x)

\int 2 cos^{4} (3 x) d x

y^{''}-4y^'+13y=sin(x)-e^{3x}+5

y^{^{''}} - 4 y^{^{'}} + 13 y = sin (x) - e^{3 x} + 5

tangent of (x^3-16x)^8 tangent of

(x^{3} - 16 x)^{8}

(\partial)/(\partial y)(2x^4y+4)

\frac{\partial}{\partial y} (2 x^{4} y + 4)

derivative of sec(0)

\frac{d}{d x} (sec (0))

Frequently Asked Questions (FAQ)

What is the solution for y^'-e^{9x+y}=0 ?
The solution for y^'-e^{9x+y}=0 is y=-ln(-1/9 e^{9x}-c_{1})