Answer to Question #93053 in Differential Equations for Subhasis Padhy

Question

Find the general solution of the differential equation

y²/2 + 2y exp(t) + (y+exp(t)) dy/dt = 0

Answer 1

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

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

$P(t,y)=\frac{y^2}{2}+2ye^t$

$Q(t,y)=y+e^t$

$∂P/∂y=y+2e^t$

$∂Q/∂t=e^t$

$∂P/∂y \not = ∂Q/∂t$, so equation is not exact.

Let's find an integrating factor $\mu$ :

$\xi=\frac{∂P/∂y - ∂Q/∂t}{Q}=\frac{y+2e^t-e^t}{y+e^t}=\frac{y+e^t}{y+e^t}=1$

$\xi$ is a function of t only (not y), let it be denoted by $\xi (t)$. Then

$\mu=e^{\int\xi(t)dt}=e^{\int dt}=e^t$

Differential equation with integrating factor:

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

$P_1(t,y)=\frac{y^2e^t}{2}+2ye^{2t}$

$Q_1(t,y)=ye^t+e^{2t}$

$∂P_1/∂y=ye^t+2e^{2t}$

$∂Q_1/∂t=ye^t+2e^{2t}$

$∂P_1/∂y = ∂Q_1/∂t$, so equation $(\frac{y^2e^t}{2}+2ye^{2t})dt+(ye^t+e^{2t})dy=0$ is exact.

So, solution of this equation $f(t,y)$ and $∂f(t,y)/∂t=P_1(t,y)$ and $∂f(t,y)/∂y=Q_1(t,y)$.

$f(t,y)=\int P_1(t,y)dt +\phi(y)$

$f(t,y)=\int (\frac{y^2e^t}{2}+2ye^{2t})dt +\phi(y)=\frac{y^2e^t}{2}+ye^{2t} +\phi(y)$

$∂f(t,y)/∂y=(\frac{y^2e^t}{2}+ye^{2t} +\phi(y))'=ye^{t}+e^{2t}+\phi'(y) =Q_1(t,y)$

$ye^{t}+e^{2t}+\phi'(y) =ye^t+e^{2t}$

$\phi'(y) =0$

$\phi(y) =C_1$

$f(t,y)=\frac{y^2e^t}{2}+ye^{2t} +C_1$

So, a solution of the differential equation is

$f(t,y)=C_2$

$\frac{y^2e^t}{2}+ye^{2t} +C_1=C_2,\, C=C_2-C_1,$

$\frac{y^2e^t}{2}+ye^{2t} =C$

Answer: $\frac{y^2e^t}{2}+ye^{2t} =C$