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Popular >

y(x-b)+y^'bx=0

Solution

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

Solution

y = e^{- \frac{x}{b} + c_{1}} x

Solution steps

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

Solve separable ODE:

y = e^{- \frac{x}{b} + c_{1}} x

y = e^{- \frac{x}{b} + c_{1}} x

Graph

Popular Examples

y^'=(-y^3+14x-8)/(3xy^2)

y^{^{'}} = \frac{- y ^{3} + 14 x - 8}{3 x y ^{2}}

derivative of 2/(2x^4-5)

\frac{d}{d x} (\frac{2}{2 x ^{4} - 5})

derivative of ((x+1^2+(x-1)^2)/(x^2))

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

integral of (2x+3)/(x^2+2x+5)

\int \frac{2 x + 3}{x ^{2} + 2 x + 5} d x

(\partial)/(\partial x)(e^{-y/(x^2+y^2+1)})

\frac{\partial}{\partial x} (e^{- \frac{y}{x ^{2} + y ^{2} + 1}})

Frequently Asked Questions (FAQ)

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