Math 217: Differential Equations
Lecture Notes: §1.2 Integrals as General and Particular
Solutions
Mark Pedigo
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Integrating to Find General and Particular Solutions
The first type of DE we look at in this course is of the form
dy
= f (x),
dx
where the right-hand side does not involve the dependent value y. In this special case, we
need only integrate both sides of the equation to obtain
y(x) =
Z
f (x) dx + C.
Since this solution involves an arbitrary constant, it is a general solution of the DE.
Example 1.1 (Integration as a General Solution). Solve:
dy
dx
= 2x + 3.
Solution. Integrating both sides yields the general solution
y(x) =
Z
(2x + 3) dx = x2 + 3x + C.
If we have an IC, we can solve for the arbitrary constant to obtain a particular solution.
So, given the IVP
dy
= f (x), y(x0 ) = y0 ,
dx
we first integrate (as above) and then solve for the arbitrary constant C using the IC.
Example 1.2 (Integration as a Particular Solution). Solve:
dy
dx
= 2x + 3, y(1) = 2.
Solution. As before, integrating both sides yields the general solution
y(x) =
Z
(2x + 3) dx = x2 + 3x + C.
By the IC, y(1) = 2, from which we see y(1) = 12 + 3 · 1 + C = 2, so C = −2. Thus, the
particular solution is
y(x) = x2 + 3x − 2.
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Recommended Homework
• §1.2, 1-10 (p. 17). Solve the given IVPs using integration.
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