The answer is
t dy / dt + dy / dt = te ^ y
I apply common factor 1/dt*(tdy+dy)=te^y
I pass "dt" tdy+dy=(te^y)*dt
I apply common factor (t+1)*dy=(te^y)*dt
I pass "e^y" (1/e^y)dy=((t+1)*t)*dt
I apply integrals ∫ (1/e^y)dy= ∫ (t^2+t)*dt
by property of integrals ∫ (1/e^y)dy= ∫ (e^-y)dy
∫ (e^-y)dy= ∫ (t^2+t)*dt
I apply integrals
-e^-y=(t^3/3)+(t^2/2)+C
I apply natural logarithm to eliminate "e"
-ln (e^-y)=-ln(t^3/3)+(t^2/2)+C
y=ln(t^3/3)+(t^2/2)+C
