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Popular >

y^'+4xy=2x

Solution

y^{^{'}} + 4 x y = 2 x

Solution

y = \frac{e ^{- 2 x^{2} + c_{1}}}{2} + \frac{1}{2}

Solution steps

y^{'} (x) + 4 x y = 2 x

Solve separable ODE:

y = \frac{e ^{- 2 x^{2} + c_{1}}}{2} + \frac{1}{2}

y = \frac{e ^{- 2 x^{2} + c_{1}}}{2} + \frac{1}{2}

Popular Examples

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

y^'+ycos(x)=3cos(x),y(0)=5

y^{^{'}} + y cos (x) = 3 cos (x), y (0) = 5

x(dy)/(dx)-7y=0

x \frac{d y}{d x} - 7 y = 0

e^yy^'=((ln(t))^2)/t ,y(1)=ln(19)

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}, y (1) = ln (19)

((dy)/(dx))/y = 3/(\sqrt[5]{2x^7+5x+1)}

Frequently Asked Questions (FAQ)

What is the solution for y^'+4xy=2x ?
The solution for y^'+4xy=2x is y=(e^{-2x^2+c_{1}})/2+1/2