What is the solution for (dy)/(dx)=4y^2-3y+1 ?

The solution for (dy)/(dx)=4y^2-3y+1 is y=(tan((sqrt(7)(x+c_{1}))/2)sqrt(7))/(8)+3/8

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(dy)/(dx)=4y^2-3y+1

\frac{d y}{d x} = 4 y^{2} - 3 y + 1

y = \frac{tan ( \frac{7 ( x + c _{1} )}{2} ) 7}{8} + \frac{3}{8}

Solution steps

\frac{d y}{d x} = 4 y^{2} - 3 y + 1

Solve separable ODE:

y = \frac{tan ( \frac{7 ( x + c _{1} )}{2} ) 7}{8} + \frac{3}{8}

y = \frac{tan ( \frac{7 ( x + c _{1} )}{2} ) 7}{8} + \frac{3}{8}

x = y + 12, x = y^{2} + 2 y

sin (tan (7 x))

\int_{2}^{4} (10 + 4 x - 3 x^{3}) d x

f (x) = 30 e^{x^{2}} x

\int_{0}^{9} 2 π y (81 - y^{2}) d y

What is the solution for (dy)/(dx)=4y^2-3y+1 ?
The solution for (dy)/(dx)=4y^2-3y+1 is y=(tan((sqrt(7)(x+c_{1}))/2)sqrt(7))/(8)+3/8