What is the solution for y^'(3x+1/y)+3y+e^x=0 ?

The solution for y^'(3x+1/y)+3y+e^x=0 is y=-(e^x)/4+(c_{1})/(e^{3x)}

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y^'(3x+1/y)+3y+e^x=0

y^{^{'}} (3 x + \frac{1}{y}) + 3 y + e^{x} = 0

y = - \frac{e ^{x}}{4} + \frac{c _{1}}{e ^{3 x}}

Solution steps

y^{'} (x) + 3 y + e^{x} = 0

Solve linear ODE:

y = - \frac{e ^{x}}{4} + \frac{c _{1}}{e ^{3 x}}

y = - \frac{e ^{x}}{4} + \frac{c _{1}}{e ^{3 x}}

\int_{0}^{1} 3 ln (4 x) d x

\frac{d y}{d x} = 9 x y

\int x (x - 2) d x

4 \frac{d ^{2} y}{d x ^{2}} - 5 y = 0

What is the solution for y^'(3x+1/y)+3y+e^x=0 ?
The solution for y^'(3x+1/y)+3y+e^x=0 is y=-(e^x)/4+(c_{1})/(e^{3x)}