Unformatted text preview: Section A.1 Linear Ordinary Diﬀerential Equations A167 Solutions to Exercises A.1
1. We solve the equation y + y = 1 in two diﬀerent ways. The ﬁrst method basically
rederives formula (2) instead of just appealing to it.
Using an integrating factor. In the notation of Theorem 1, we have p(x) = 1
and q(x) = 1. An antiderivative of p(x) is thus 1 · dx = x. The integrating factor
is
µ(x) = e p(x) dx = ex .
Multiplying both sides of the equation by the integrating factor, we obtain the
equivalent equation
ex [y + y] = ex ;
dx
[e y] = ex ,
dx
d
where we have used the product rule for diﬀerentiation to set dx [exy] = ex [y + y].
Integrating both sides of the equation gets rid of the derivative on the left side, and
on the right side we obtain ex dx = ex + C . Thus, ex y = ex + C y = 1 + Ce−x, ⇒ where the last equality follows by multiplying by e−x the previous equality. This
gives the solution y = 1 + Ce−x up to one arbitrary constant, as expected from the
solution of a ﬁrst order diﬀerential equation.
Using formula (2). We have, with p(x) = 1, p(x) dx = x (note how we took
the constant of integration equal 0):
y = e−x C + 1 · ex dx = e−x [C + ex ] = 1 + Ce−x . 5. According to (2),
y = ex C + sin xe−x dx . To evaluate the integral, use integration by parts twice
sin xe−x dx =
= − sin xe−x + e−x cos x dx − sin xe−x + cos x(−e−x ) − So sin xe−x dx = −e−x sin x + cos x sin xe−x dx = 2 e−x sin x dx; 1
− e−x sin x + cos x .
2 1
y = ex C − e−x sin x + cos x
2 = Cex − 1
sin x + cos x .
2 9. We use an integrating factor
e p(x) dx =e tan x dx = e− ln(cos x) = 1
= sec x.
cos x Then
sec xy − sec x tan x y = d
[y sec x] =
dx
y sec x =
y = sec x cos x;
1
x+C =
x cos x + C cos x. ...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk

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