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