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

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

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

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

y = arctan (t) (1 + t^{2}) + t + c_{1} (1 + t^{2})

Solution steps

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

Solve linear ODE:

y = arctan (t) (1 + t^{2}) + t + c_{1} (1 + t^{2})

y = arctan (t) (1 + t^{2}) + t + c_{1} (1 + t^{2})

\frac{d}{d x} (9 - 2 x)

f (x) = x^{2} - 6 x + 1, at x = 1

f (x) = \frac{1}{ln ( x )}

x \to 5 lim (\frac{\frac{5}{x} - 1}{x - 5})

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