What is the solution for (t^2+1)f^'(t)+[f(t)]^2+1=0,f(4)=e ?

The solution for (t^2+1)f^'(t)+[f(t)]^2+1=0,f(4)=e is f(t)=tan(-arctan(t)+1.21828…+1.32581…)

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(t^2+1)f^'(t)+[f(t)]^2+1=0,f(4)=e

(t^{2} + 1) f^{^{'}} (t) + [f (t)]^{2} + 1 = 0, f (4) = e

f (t) = tan (- arctan (t) + 1.21828 \dots + 1.32581 \dots)

Solution steps

(t^{2} + 1) f^{'} (t) + [f (t)]^{2} + 1 = 0

Solve separable ODE:

f (t) = tan (- arctan (t) + 1.21828 \dots + 1.32581 \dots)

f (t) = tan (- arctan (t) + 1.21828 \dots + 1.32581 \dots)

\int \frac{x ^{3}}{x} d x

\int_{0}^{2} 5 x^{3} x^{2} + 4 d x

x \to 3 - lim (c x^{2} + 4 x)

\int \frac{x ^{2} + 5 x + 1}{( x ^{2} + 1 ) ^{2}} d x

\frac{- 1}{( 1 + x ) ^{2}}

What is the solution for (t^2+1)f^'(t)+[f(t)]^2+1=0,f(4)=e ?
The solution for (t^2+1)f^'(t)+[f(t)]^2+1=0,f(4)=e is f(t)=tan(-arctan(t)+1.21828…+1.32581…)