What is the solution for (2x+1)y^'+2y=4x ?

The solution for (2x+1)y^'+2y=4x is y=(2x^2)/(2x+1)+(c_{1})/(2x+1)

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(2x+1)y^'+2y=4x

(2 x + 1) y^{^{'}} + 2 y = 4 x

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

Solution steps

(2 x + 1) y^{'} (x) + 2 y = 4 x

Solve linear ODE:

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

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

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

\int \frac{1}{5 y} d y

y^{^{''}} + y = 0, y (\frac{π}{3}) = 0, y^{^{'}} (\frac{π}{3}) = 8

\frac{\partial}{\partial x} (3 x + 9 x^{17})

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