Answer:[tex]y^2=2\ln (1+t^2)+4[/tex]Step-by-step explanation:Given that[tex]\dfrac{dy}{dt}=\dfrac{2t}{y+yt^2}[/tex]This is a differential equation.Now by separating variables[tex]y dy= \dfrac{2t}{1+t^2}dt[/tex]Now by integrating both side[tex]\int y dy=\int \frac{2t}{1+t^2}dt[/tex]Now by soling above integrationWe know that integration of dx/x is lnx.[tex]\dfrac{y^2}{2}=\ln (1+t^2)+C[/tex]Where C is the constant.[tex]y^2=2\ln (1+t^2)+C[/tex]Given that when t=0 then y= -2So by putting the above values of t and y we will find C4=2 ln(1)+C (we know that ln(1)=0)So C=4⇒[tex]y^2=2\ln (1+t^2)+4[/tex]So solution of above equation is [tex]y^2=2\ln (1+t^2)+4[/tex]