Question

Find the general solution of the differential equation .
y²/2 + 2yexp(t) + (y+exp(t))dy/dt = 0

Answer 1

$y^2/2+2ye^t+(y+e^t)y'=0$

Multiply by 2

$y^2+4ye^t+2(y+e^t)y'=0$

$y^2+2ye^t+e^{2t}+2(y+e^t)y'+2ye^t+2e^{2t}-3e^{2t}=0$

$(y+e^t)^2+2(y+e^t)y'+2(y+e^t)e^t-3e^{2t}=0$

$(y+e^t)^2+2(y+e^t)(y'+e^t)=3e^{2t}$

Notice that

$((y+e^t)^2)'=2(y+e^t)(y'+e^t)$

Let $z=(y+e^t)^2$

then

$z+z'=3e^{2t}$

Using an integrating factor

$u(t)=e^{\int1dt}=e^t$

General solution is

$z=\frac{\intop e^t3e^{2t}dt+C}{e^t}=\frac{e^{3t}+C}{e^t}$

$(y+e^t)^2=\frac{e^{3t}+C}{e^t}$

$y+e^t=\pm \sqrt{\frac{e^{3t}+C}{e^t}}$

$y=\pm \sqrt{\frac{e^{3t}+C}{e^t}}-e^t$

Answer: $y=\pm \sqrt{\frac{e^{3t}+C}{e^t}}-e^t$