Answer
$y = \dfrac{\tan^{-1}t + C}{(1 + t^2)^2}$
Work Step by Step
$$(1 + t^2)y' + 4ty = (1 + t^2)^{-2}$$
Divide through by ((1 + t^2)):
$$y' + \frac{4t}{1 + t^2}y = (1 + t^2)^{-3}$$
We solve this by the integrating factor method:
$$\mu(t) = e^{\int P(t),dt}$$
Here, (P(t) = \frac{4t}{1 + t^2}).
Therefore,
$$\mu(t) = e^{\int \frac{4t}{1 + t^2},dt} = e^{2\ln(1 + t^2)} = (1 + t^2)^2$$
Now, multiplying both sides by (\mu(t)):
$$(1 + t^2)^2(y' + \frac{4t}{1 + t^2}y) = (1 + t^2)^2(1 + t^2)^{-3}$$
$$(1 + t^2)^2y' + 4t(1 + t^2)y = (1 + t^2)^{-1}$$
Notice the left-hand side:
$$((1 + t^2)^2y)' = (1 + t^2)^{-1}$$
Integrating both sides:
$$\int ((1 + t^2)^2y)',dt = \int (1 + t^2)^{-1},dt$$
On the right-hand side:
$$\int (1 + t^2)^{-1},dt = \tan^{-1}t + C$$
Therefore:
$$(1 + t^2)^2y = \tan^{-1}t + C$$
Dividing both sides by ((1 + t^2)^2):
$$y = \frac{\tan^{-1}t + C}{(1 + t^2)^2}$$
