What is the solution for (dx)/(dt)=6(x^2+1),x(pi/4)=1 ?

The solution for (dx)/(dt)=6(x^2+1),x(pi/4)=1 is x=tan(6t-(5pi)/4)

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(dx)/(dt)=6(x^2+1),x(pi/4)=1

\frac{d x}{d t} = 6 (x^{2} + 1), x (\frac{π}{4}) = 1

x = tan (6 t - \frac{5 π}{4})

Solution steps

\frac{d x}{d t} = 6 (x^{2} + 1)

Solve separable ODE:

x = tan (6 t - \frac{5 π}{4})

x = tan (6 t - \frac{5 π}{4})

\frac{d ^{2}}{d x ^{2}} (\frac{x}{x ^{2} - 4})

\frac{d}{d x} (4 ln ((\frac{x}{4})^{2} - 1))

x \to 0 + lim (2 + x^{9} ln (x))

\frac{\partial}{\partial y} (x y e^{yz})

\int 9 e^{u} d u

What is the solution for (dx)/(dt)=6(x^2+1),x(pi/4)=1 ?
The solution for (dx)/(dt)=6(x^2+1),x(pi/4)=1 is x=tan(6t-(5pi)/4)