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

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

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

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

y = ln (2 x + 1) (2 x + 1) + 1 + c_{1} (2 x + 1)

Solution steps

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

Solve linear ODE:

y = ln (2 x + 1) (2 x + 1) + 1 + c_{1} (2 x + 1)

y = ln (2 x + 1) (2 x + 1) + 1 + c_{1} (2 x + 1)

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