What is the solution for y^'=(3^2y^2+1)/3 ,y(0)=-0.5 ?

The solution for y^'=(3^2y^2+1)/3 ,y(0)=-0.5 is y=(tan(t-0.98279…))/3

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y^'=(3^2y^2+1)/3 ,y(0)=-0.5

y^{^{'}} = \frac{3 ^{2} y ^{2} + 1}{3}, y (0) = - 0.5

y = \frac{tan ( t - 0.98279 \dots )}{3}

Solution steps

y^{'} (t) = \frac{3 ^{2} y ^{2} + 1}{3}

Solve separable ODE:

y = \frac{tan ( t - 0.98279 \dots )}{3}

y = \frac{tan ( t - 0.98279 \dots )}{3}

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What is the solution for y^'=(3^2y^2+1)/3 ,y(0)=-0.5 ?
The solution for y^'=(3^2y^2+1)/3 ,y(0)=-0.5 is y=(tan(t-0.98279…))/3