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24 1. First-Order Differential Equations
then x(2) = ± √ 2 2 + C = 1, then we are forced to take the positive square root
and obtain C = −3. Therefore, the solution to the IVP
is
dx
dt = t x , x(2) = 1
x(t) = √ t 2 − 3.
The solution to the IVP is defined only when t > √ 3, the interval of
existence. □
Remark 1.15
(Recipe) The method of separation of variables for the equation
just results in writing down
dx
dt = f(x)g(t)
1
dx = g(t) dt,
f(x)
where the x terms are placed on the left and the t terms on the right, including
the two differentials dx and dt. Then we integrate to get
∫
∫
1
f(x) dx = g(t)dt + C.
This is the recipe used to solve problems. We usually dispense with integrating
both sides with respect to t and then changing variables. We go directly to the
last equation. □
Example 1.16
(Growth and Decay) Consider the differential equation
dx
= rx, (1.13)
dt
where r is a given constant. If r<0 this is the decay equation; if r>0thenit
models exponential growth. This equation is separable. We write
1
dx = rdt.
x
Integrating gives ∫ ∫ 1
x dx = r dt,