What is the solution for (2x+1)y^'+2y=4x,y(0)=1 ?

The solution for (2x+1)y^'+2y=4x,y(0)=1 is y=(2x^2)/(2x+1)+1/(2x+1)

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(2x+1)y^'+2y=4x,y(0)=1

(2 x + 1) y^{^{'}} + 2 y = 4 x, y (0) = 1

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

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

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25 y^{^{''}} - y = x e^{\frac{x}{5}}, y (0) = 1, y^{^{'}} (0) = 0

x \to \infty lim (\frac{1}{x ( ln ( x ) ) ^{4}})

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What is the solution for (2x+1)y^'+2y=4x,y(0)=1 ?
The solution for (2x+1)y^'+2y=4x,y(0)=1 is y=(2x^2)/(2x+1)+1/(2x+1)