Toronto Math Forum

Find the general solution of the given differential equation.
$$ y'' + 2y' + 2y = 0$$

Solve the equation $r^2+2r+2=0$ for $r$
$$ r = \frac{-2 \pm \sqrt{2^2-4(1)(2)} }{2(1)} $$
$$ = \frac{-2 \pm \sqrt{-4}}{2} = \frac {-2 \pm 2i}{2} = -1 \pm i$$

We know that $y = c_1e^{\lambda t}cos(\mu t) + c_2e^{\lambda t}sin(\mu t)$ and $\lambda = -1$ and $\mu = 1$

Therefore, the required general solution is
$$ y = c_1e^{-t}cos(t) + c_2e^{-t}sin(t) $$

Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$x^2y^3 + x(1+y^2)y' = 0,   ~~~~  \mu(x, y) = \frac{1}{xy^3}$$
First, let's show the given equation isn not exact.
Define $M(x, y) = x^2y^3, ~~ N(x, y) =  x(1+y^2)$
$$M_y = \frac{\partial}{\partial y}[x^2y^3] = 3x^2y^2$$
$$N_x = \frac{\partial}{\partial x}[x(1+y^2)] = 1 + y^2$$
Since $M_y \neq N_x$, this implies that the given equation is not exact.

Next, show that the given equation multiplied by the integrating factor $\mu(x, y) = \frac{1}{xy^3}$ is exact.
The new equation becomes
$$ \frac{1}{xy^3}x^2y^3 + \frac{1}{xy^3}x(1+y^2)y' = x + (y^{-3} + y^{-1})y' =0 $$
Define $M'(x, y) = x, ~~ N'(x, y) =  y^{-3} + y^{-1}$
$$M'_y = \frac{\partial}{\partial y}(x) = 0$$
$$N'_x = \frac{\partial}{\partial x}(y^{-3} + y^{-1}) = 0$$
Since $M'_y = N'_x$, this implies that the given equation is exact.

Thus, we know that there exists a function $\phi (x, y) = C$ which satisfies the give differential equation.
Also,
$$\frac{\partial \phi}{\partial x} = M'(x, y) = x$$
$$\frac{\partial \phi}{\partial y} = N'(x, y) = y^{-3} + y^{-1}$$
Integrate $\frac{\partial \phi}{\partial x} = x$ with respect to $x$ we have
$$\phi (x, y) = \frac{1}{2}x^2 + h(y)$$
Take derivative on both sides with respect to $y$ we get
$$\frac{\partial \phi}{\partial y} = h'(y)$$
Since we know $\frac{\partial \phi}{\partial y} = N'(x, y) = y^{-3} + y^{-1}$
Then $$ h'(y) = y^{-3} + y^{-1}$$
Integrate $h'(y)$ with respect to y we have
$$ h(y) = -\frac{1}{2}y^{-2} + ln|y| + C$$
Therefore, we have
$$ \phi (x, y) = \frac{1}{2}x^{2} - \frac{1}{2}y^{-2} + ln|y| = C$$
is the general solution to the given differential equation.