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Popular >

(8x-1)y^'+8y=0

Solution

(8 x - 1) y^{^{'}} + 8 y = 0

Solution

y = \frac{c _{1}}{8 x - 1}

Solution steps

(8 x - 1) y^{'} (x) + 8 y = 0

Solve separable ODE:

y = \frac{c _{1}}{8 x - 1}

y = \frac{c _{1}}{8 x - 1}

Popular Examples

cos(y)dx+e^{-x}sin(y)dy=0

cos (y) d x + e^{- x} sin (y) d y = 0

d/(df)(\H(f))=(1+u(f)^2)/(1+f^2)

\frac{d}{df} (H (f)) = \frac{1 + u ( f ) ^{2}}{1 + f ^{2}}

d y - 2 y d x = 4 d x

3y^2(xdy+ydx)=ydx-xdy

3 y^{2} (x d y + y d x) = y d x - x d y

(dy)/(dt)+4y=11

\frac{d y}{d t} + 4 y = 11

Frequently Asked Questions (FAQ)

What is the solution for (8x-1)y^'+8y=0 ?
The solution for (8x-1)y^'+8y=0 is y=(c_{1})/(8x-1)