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7.4. SEPARABLE DIFFERENTIAL EQUATIONS
we know that the solution has the form y(t) =
3
3
2 t 2 + C for some constant C. We
therefore must find the appropriate value for C that gives the initial value y(0) = 2. Hence,
2 = y(0) =
3
3
2
0 2 + C =
3
√
C,
which shows that C = 2 3 = 8. The solution to the initial value problem is then
y(t) =
3
3
2
t 2 + 8.
The strategy of Example 7.1 may be applied to any differential equation of the form
dy
dt = g(y) · h(t), and any differential equation of this form is said to be separable. We work
to solve a separable differential equation by writing
1
g(y)
dy
dt
= h(t),
and then integrating both sides with respect to t. After integrating, we strive to solve
algebraically for y in order to write y as a function of t.
We consider one more example before doing further exploration in some activities.
Example 7.2. Solve the differential equation
dy
dt
= 3y.
Solution. Following the same strategy as in Example 7.1, we have
1
y
dy
dt
= 3.
Integrating both sides with respect to t,
1
y
dy
dt
dt =
3 dt,
and thus
1
y
dy =
3 dt.
Antidifferentiating and including the integration constant, we find that
ln |y| = 3t + C.