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Summary: first order differential equations
Types discussed in class.
- Separable equations. These are equations which may be written in the form y′^ = f (y)g(t). To solve, you separate the variables:
1 f (y) dy = g(t)dt.
Then integrate, making sure to include one of the constants of integration: ∫ 1 f (y)
dy =
g(t)dt + c.
- Linear equations. These are equations of this form:
y′^ + p(t)y = q(t).
To solve, make sure it’s in exactly this form, and multiply by the “inte- grating factor” e
p(t)dt:
e
p(t)dty′ (^) + e
p(t)dtp(t)y = e
p(t)dtq(t).
Let I(t) denote the integrating factor: I(t) = e
p(t)dt. Then the left side of the equation is the derivative of I(t)y:
(I(t)y)′^ = I(t)q(t).
Now integrate: I(t)y =
I(t)q(t)dt + c,
so y =
I(t)
I(t)q(t)dt +
c I(t)
(I find it much easier to remember the procedure—multiply by e
p(t)dt— than to try to memorize this solution.)
Types not discussed in class. You won’t need to know these for any home- work or exams for this class, but they might come up in other courses.
- Homogeneous equations. (See Section 2.9.) (Warning: the word “ho- mogeneous” gets used in several different ways when studying differential equations.) A homogeneous equation is one of this form:
y′^ = f (x, y),
where the function f (x, y) depends only on the ratio y/x—say f (x, y) = (y/x)^2 + sin(3 y/x). In other words, it is an equation of the form
y′^ = F (y/x)
for some function F. Let v = y/x, so that y = vx. Then y′^ = v′x + v, and if you plug this in for y′^ and plug v in for y/x, you get
xv′^ + v = F (v).
This is separable: dv F (v) − v
dx x
So solve it for v, and then substitute back in for y: v = y/x.
- Bernoulli equations. (See problems 37–41 in Section 2.2.) These are a lot like linear equations; they are equations of this form:
y′^ + p(t)y = q(t)yn,
where n is any number except 0 or 1. (If n = 0, then y^0 = 1, so this is just a linear equation. If n = 1, then y^1 = y, so you can rewrite this as y′^ + [p(t) − q(t)]y = 0, which is a linear equation.) To solve it, make the substitution v = y^1 −n, so that v′^ = (1 − n)y−ny′; in other words, y−ny′^ = (^1) −^1 n v′. Multiply the original equation by y−n:
y−ny′^ + p(t)y^1 −n^ = q(t).
Now make the substitution with v: 1 1 − n
v′^ + p(t)v = q(t).
Multiply everything by 1 − n and you have a linear equation, which you can solve to find v. Once you have v, then use the equation y = v^1 /(1−n) to find y.
