What is the solution for (dx)/(dt)=x^2+1/49 ,x(0)=1 ?

The solution for (dx)/(dt)=x^2+1/49 ,x(0)=1 is x=(tan((t+7*1.42889…)/7))/(7)

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(dx)/(dt)=x^2+1/49 ,x(0)=1

\frac{d x}{d t} = x^{2} + \frac{1}{49}, x (0) = 1

x = \frac{tan ( \frac{t + 7 \cdot 1.42889 \dots}{7} )}{7}

Solution steps

\frac{d x}{d t} = x^{2} + \frac{1}{49}

Solve separable ODE:

x = \frac{tan ( \frac{t + 7 \cdot 1.42889 \dots}{7} )}{7}

x = \frac{tan ( \frac{t + 7 \cdot 1.42889 \dots}{7} )}{7}

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0, y (0) = 6

ln (x + 1)

\frac{d ^{3}}{d x ^{3}} (sin (7 x))

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (1) = 0, y^{^{'}} (1) = 1

f (x) = 3 - x - x^{2}, at x = 1

What is the solution for (dx)/(dt)=x^2+1/49 ,x(0)=1 ?
The solution for (dx)/(dt)=x^2+1/49 ,x(0)=1 is x=(tan((t+7*1.42889…)/7))/(7)