Step-by-step explanation:
[tex] \frac{dy}{dt} = \frac{t}{ {t}^{2} y + y} [/tex]
[tex] \frac{dy}{dt} = \frac{t}{y( {t}^{2} + 1) } [/tex]
Using seperation by variables,
[tex]ydy = \frac{t}{ {t}^{2} + 1 } dt[/tex]
Integrate both sides, for the left side, use the power rule, we get
[tex] \frac{y {}^{2} }{2} [/tex]
For the right side, let
[tex]u = {t}^{2} + 1[/tex]
[tex]du = 2tdt[/tex]
[tex] \frac{du}{2} = tdt[/tex]
So we get, on the right side
[tex] \frac{1}{2} ln( |t {}^{2} + 1 | ) + c[/tex]
As of right now, we have
[tex] \frac{ {y}^{2} }{2} = \frac{1}{2} ln( {t}^{2} + 1) + c[/tex]
Solving for y, gives us
[tex] {y}^{2} = ln( {t}^{2} + 1 ) + c[/tex]
[tex]y = \sqrt{ ln( {t}^{2} + 1) + c} [/tex]
Plugging in t=0, gives us
[tex]2 = \sqrt{ ln(1) + c } [/tex]
[tex]2 = \sqrt{c} [/tex]
[tex]4 = c[/tex]
So. our formula is
[tex]y = \sqrt{ ln( {t}^{2} + 1 ) + 4 } [/tex]
