Answer to Question #95327 in Differential Equations for Pooja panghal

Question

Find the general solution of the differential eq. y^2/2+2ye^t+(y+e^t)dy/dt=0

Answer 1

let's rewrite the equation in the form

$(\frac{y^2}{2}+2ye^t)dt+(y+e^t)dy=0$

This is not an exact equation, because

$\frac{\partial}{\partial y}(\frac{y^2}{2}+2ye^t)=y+2e^t\ne e^t=\frac{\partial}{\partial t}(y+e^t)$

Therefore we need to find an integrating factor $\mu$ , such that

$\frac{\partial}{\partial y}\mu(t)(\frac{y^2}{2}+2ye^t)=\frac{\partial}{\partial t}\mu(t)(y+e^t)$

$\mu(t)(y+2e^t)=\frac{d\mu}{dt}(y+e^t)+\mu(t)e^t$

$\mu(t)(y+e^t)=\frac{d\mu}{dt}(y+e^t)$

$\mu(t)=\frac{d\mu}{dt}$

$\mu=e^t$

Then

$(e^t\frac{y^2}{2}+2ye^{2t})dt+(e^ty+e^{2t})dy=0$

is an exact equation

Let's integrate

$\int(e^t\frac{y^2}{2}+2ye^{2t})dt=e^t\frac{y^2}{2}+4ye^{2t}+C(y)$

Then

$\frac{d}{dy}\left(e^t\frac{y^2}{2}+4ye^{2t}+C(y)\right)\\=ye^t+4e^{2t}+C'=ye^t+e^{2t}$

$C'=-3e^{2t}$

$C=-\frac{3e^{2t}}{2}$

Therefore, the solution of the equation will be

$e^t\frac{y^2}{2}+4ye^{2t}-\frac{3e^{2t}}{2}=const$