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**Unformatted text preview: **Math 54 quiz 9 Solutions
1. Find the general solution to the diﬀerential equation
y + 2y + ky = 0,
where (a) k = 0, (b) k = 1, and (c) k = 2.
(a) The roots of the characteristic equation are 0, and -2, so the solution
is
y (x) = c1 + c2 e−2x .
(b) The root −1 is a repeated root, so the solution is
y (x) = c1 e−x + c2 xe−x .
(c) The (complex) roots are −1 ± i so the solution is
y (x) = e−x (c1 sin(x) + c2 cos(x)).
2. Suppose that y1 satisﬁes the diﬀerential equation
x2 y + sin(x)y = f (x),
and that y2 satisﬁes the diﬀerential equation
x2 y + sin(x)y = g (x).
Show that c1 y1 + c2 y2 satisﬁes
x2 y + sin(x)y = c1 f (x) + c2 g (x). In order to show that c1 y1 + c2 y2 satisﬁes the diﬀerential equation, we
need to plug it in to the left hand side and see what we get: substituting
c1 y1 + c2 y2 in for y , we obtain x2 (c1 y1 +c2 y2 ) +sin x(c1 y1 +c2 y2 ) = c1 x2 y1 +c2 x2 y2 +c1 sin(x)y1 +c2 sin(x)y2 = c1 x2 y1 + sin(x)y1 +
and recalling that y1 and y2 satisfy the diﬀerential equations given above,
this is
c1 f (x) + c2 g (x),
as required. 1 3. Find the general solution to the diﬀerential equation
y − y − 2y = e2x
We use undetermined coeﬃcients. The roots of the characteristic equation
are −2, +1, so the homogeneous solution is
yH (x) = c1 e−x + c2 e2x .
The right side is e2x , so our ﬁrst guess is Ae2x , but this is already the
second solution to the homogeneous equation, so we change the guess to
Axe2x . We take two derivatives of this guess to get Ae2x + 2Axe2x and
4Ae2x + 4Axe2x , respectively. Thus, this guess plugged into the left hand
side of the equation is
4Ae2x +4Axe2x −Ae2x −2Axe2x −2Axe2x = xe2x (4A−2A−2A)+e2x (4A−A) = 3Ae2x .
For this to be a solution, it must equal e2x , so we must have A = 1/3.
The solution then is
1
y (x) = c1 e−x + c2 e2x + xe2x
3 2 ...

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