Answer :
This ODE is separable:
[tex]\dfrac{\mathrm du}{\mathrm dt}=\dfrac{2t+\sec^2t}{2u}[/tex]
[tex]\implies2u\,\mathrm du=(2t+\sec^2t)\,\mathrm dt[/tex]
Integrating both sides gives
[tex]u^2=t^2+\tan t+C[/tex]
Given that [tex]u(0)=-5[/tex], we have
[tex](-5)^2=0^2+\tan0+C\implies C=25[/tex]
so that the particular solution to the IVP is
[tex]u(t)^2=t^2+\tan t+25[/tex]
[tex]\boxed{u(t)=\sqrt{t^2+\tan t+25}}[/tex]