What is the solution for (dy)/(dx)=1-(10x-y+4)^2 ?

The solution for (dy)/(dx)=1-(10x-y+4)^2 is y=-3tan(3x+c_{1})+10x+4

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(dy)/(dx)=1-(10x-y+4)^2

\frac{d y}{d x} = 1 - (10 x - y + 4)^{2}

y = - 3 tan (3 x + c_{1}) + 10 x + 4

Solution steps

\frac{d y}{d x} = 1 - (10 x - y + 4)^{2}

Solve using substitution:

y = - 3 tan (3 x + c_{1}) + 10 x + 4

y = - 3 tan (3 x + c_{1}) + 10 x + 4

3 y^{^{''}} + 3 y^{^{'}} - 2 y = 0

e^{(x^{2})}

x \to 1 + lim (\frac{2 x + 1}{x - 1})

y^{^{''}} + 12 y^{^{'}} + 37 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int_{0}^{1} \frac{23}{4 y - 1} d y

What is the solution for (dy)/(dx)=1-(10x-y+4)^2 ?
The solution for (dy)/(dx)=1-(10x-y+4)^2 is y=-3tan(3x+c_{1})+10x+4