What is the solution for (2x+3)y^'=y+(2x+3)^{1/2} ?

The solution for (2x+3)y^'=y+(2x+3)^{1/2} is y= 1/2 ln(2x+3)(2x+3)^{1/2}+c_{1}(2x+3)^{1/2}

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(2x+3)y^'=y+(2x+3)^{1/2}

(2 x + 3) y^{^{'}} = y + (2 x + 3)^{\frac{1}{2}}

y = \frac{1}{2} ln (2 x + 3) (2 x + 3)^{\frac{1}{2}} + c_{1} (2 x + 3)^{\frac{1}{2}}

Solution steps

(2 x + 3) y^{'} (x) = y + (2 x + 3)^{\frac{1}{2}}

Solve linear ODE:

y = \frac{1}{2} ln (2 x + 3) (2 x + 3)^{\frac{1}{2}} + c_{1} (2 x + 3)^{\frac{1}{2}}

y = \frac{1}{2} ln (2 x + 3) (2 x + 3)^{\frac{1}{2}} + c_{1} (2 x + 3)^{\frac{1}{2}}

\frac{\partial}{\partial x} (e^{- 2 y} sin (2 x))

\int \frac{9 ( x - 9 )}{( x + 5 ) ( x - 2 )} d x

\frac{d}{d x} ((\frac{4 x ^{2}}{2 - x})^{3})

\int_{0}^{2} (2 - x)^{2} d x

cos (x)

What is the solution for (2x+3)y^'=y+(2x+3)^{1/2} ?
The solution for (2x+3)y^'=y+(2x+3)^{1/2} is y= 1/2 ln(2x+3)(2x+3)^{1/2}+c_{1}(2x+3)^{1/2}