What is the solution for y^'=cos(x)(3y-1) ?

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

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y^'=cos(x)(3y-1)

y^{^{'}} = cos (x) (3 y - 1)

y = \frac{e ^{3 s i n (x) + c_{1}}}{3} + \frac{1}{3}

Solution steps

y^{'} (x) = cos (x) (3 y - 1)

Solve separable ODE:

y = \frac{e ^{3 s i n (x) + c_{1}}}{3} + \frac{1}{3}

y = \frac{e ^{3 s i n (x) + c_{1}}}{3} + \frac{1}{3}

ln (\frac{d y}{d x}) = a x + b y

2 y y^{^{'}} = 15 t^{2}, y (0) = 4

\frac{d y}{d x} = - \frac{7}{4} x^{3} y^{3}

x^{2} d y = (9 + y^{2}) d x

(1 - y^{2})^{\frac{1}{2}} d x - (1 - x^{2})^{\frac{1}{2}} d y = 0

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