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Popular >

y^'+5xy=x

Solution

y^{^{'}} + 5 x y = x

Solution

y = \frac{e ^{\frac{- 5 x ^{2} + c _{1}}{2}}}{5} + \frac{1}{5}

Solution steps

y^{'} (x) + 5 x y = x

Solve separable ODE:

y = \frac{e ^{\frac{- 5 x ^{2} + c _{1}}{2}}}{5} + \frac{1}{5}

y = \frac{e ^{\frac{- 5 x ^{2} + c _{1}}{2}}}{5} + \frac{1}{5}

Graph

Popular Examples

(dy)/(dx)=((ln(x))^6)/(xy^2),y(1)=4

\frac{d y}{d x} = \frac{( ln ( x ) ) ^{6}}{x y ^{2}}, y (1) = 4

y^'=((ln(x))^3)/(xy^2)

y^{^{'}} = \frac{( ln ( x ) ) ^{3}}{x y ^{2}}

derivative of (y^2)+(dy)/(dx)-2y=0

\frac{d}{d x} ((y)^{2}) + \frac{d y}{d x} - 2 y = 0

y^{^{'}} + x y = 5 x y^{2}

\frac{x}{y} = \frac{d y}{d x}

Frequently Asked Questions (FAQ)

What is the solution for y^'+5xy=x ?
The solution for y^'+5xy=x is y=(e^{(-5x^2+c_{1})/2})/(5)+1/5