Solve the differential equation t dy/dt + dy/dt = te ^y

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

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