What is the solution for (1+t^2)(dy)/(dt)=1+y^2 ?

The solution for (1+t^2)(dy)/(dt)=1+y^2 is y=tan(arctan(t)+c_{1})

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(1+t^2)(dy)/(dt)=1+y^2

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

y = tan (arctan (t) + c_{1})

Solution steps

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

Solve separable ODE:

y = tan (arctan (t) + c_{1})

y = tan (arctan (t) + c_{1})

y^{^{'}} = \frac{x + x y ^{2}}{4 y}

\frac{d x}{d t} = 30 - \frac{x}{100}

y \frac{d y}{d x} = \frac{x - e ^{x}}{e ^{y^{2}}}

\frac{d y}{d x} = \frac{20 x ^{4}}{y}

\frac{d y}{d x} = 0.2 x y

What is the solution for (1+t^2)(dy)/(dt)=1+y^2 ?
The solution for (1+t^2)(dy)/(dt)=1+y^2 is y=tan(arctan(t)+c_{1})