LINEAR FIRST ORDER EQUATIONS

General Problem Form

$$\frac{dy}{dt} + p(t)y = g(t)$$

.

Qualitative Solutions:

expect a family of solutions
depending on the initial value $y(t_0)$.
Direction field can be used for qualitative analysis.

Special Equations

• $y' = 0$: has solution $y(t) = y(t_0)$.
• $y' = g(t)$: has solution $y(t) =y(t_0)+\displaystyle\int _{t_0}^t g(t)dt$.
• $y' = ry + k$: has solution $y(t) = ce^{rt} - \frac{k}{r}$.
  c? $y(t_0) = ce^{rt_0}-k/r$, so $c = (\ y(t_0)+k/r\ )/e^{rt_0}$.

Equilibrium solutions:

when $y' = 0.$ E.g. $y' + ry = k$;
if $y(t_0)=-k/r$, then $y(t) = -k/r$ for $t > t_0$.

Integrating Factor Method for $y'+p(t)y = g(t)$

• find integrating factor $\mu(t)$ so that $\mu'(t) = p(t)\mu(t)$,
  which has solution $\mu(t) = exp(\displaystyle\int p(t)dt)$;
• then $( \mu(t)y(t) )' = \mu(t)y'(t) + \mu(t)p(t)y(t) = \mu(t)g(t)$;
• so $\mu(t)y(t) = c + \displaystyle\int \mu(t)g(t)dt$.

General Solution

$$y(t) = (\ c + \int_{t_0}^t\mu(t)g(t)dt\ )/\mu(t).$$

c? $y(t_0) = c/\mu(t_0)$, so $c = y(t_0)\mu(t_0)$.

SEPARABLE EQUATIONS

Separable Equations:

assume a nonlinear $y' = f(x,y)$.

• Suppose $y' = f(x,y)$ can be rewritten as
  
  $$M(x) + N(y)\frac{dy}{dx}= 0,$$
  
  
  
  so that the variables are separated.
• The separable equation $M(x)dx + N(y)dy = 0$,
  can be solved by integration.
• $$\int N(y)dy + \displaystyle\int M(x)dx = C$$, is an equation where
  $y$ is given (usually implicitly) in terms of $x$.
• For each $C$ there is an integral curve for a solution.