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

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

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

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

y = - 3 e^{\frac{x}{3}} + c_{1} e^{\frac{x}{2}}

Solution steps

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

Solve linear ODE:

y = - 3 e^{\frac{x}{3}} + c_{1} e^{\frac{x}{2}}

y = - 3 e^{\frac{x}{3}} + c_{1} e^{\frac{x}{2}}

\frac{d}{d x} (- 4 sin (8 x))

x \to 0 lim (\frac{x}{1 + x})

x^{2} - 6, 3

f (x) = - 2 x^{4} + x^{- 2} - 3 x^{\frac{3}{4}}

cos (π - 7 x)

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