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**Unformatted text preview: **Section 2.4. Linear Equations 1. Compare y’ = —y + 2 with y’ = a(t)y + f(t) and
note that a(t) = -—1. Consequently, an integrating
factor is found with efldt_ r 140‘) = f_“m‘i’: = e . Multiply both sides of our equation by this integrat—
ing factor and check that the left-hand side of the
resulting equation is the derivative of a product. e'(y’ + y) = 2e‘
(e’y)’ = 28‘
Integrate and solve for y.
e’ y = 2e‘ + C
y(t) = 2 + C e“’ We have a(t) = 3, so u(t) = e"3’ . Multiplying we
see that the equation becomes 3—3:), I __=3e—3:y 5e‘3‘. We verify that the left-hand side is the (hi-'2‘
(2—3‘ y, so when we integrate we get 5
e-3'yo) = ~3e—3‘ + C. Solving for y, we get y(t) = —§ + Ce3'. Compare y’ + (2/x)y = (cosx)/Jt2 w'ﬁ f =
a(x)x + f(x) and note that a(x) = —2/.r_ CI.-
sequently, an integrating factor is found will —=.IxI’z’ Multiply both sides of our equation by the ﬁll“
factor and note that the left-hand side of the m “()0 = ef—a(x)dx = efZ/xdx = e21n|.r|_ 6. If we write the equations as x’ = (4/!)x + t3, we
see that a(t) = 4/t. Thus the integrating factor is “0) = e—f(4/t)dt = e—4lnr : [—4.
Multiplying by u, the equation becomes
t’4x’ — 41‘5x = 1"]. After verifying that the left~hand side is the deriva-
tive of t‘4x, we can integrate and get z-4x(t) =1nt + C.
Hence the general solution is x(t) = t4lnr + Cr“. 14. Compare y’ = y + he“ with y’ = a(x)y + f(x)
andnote thata(x) = 1. Consequently, an integrating
factor is found with “(/0 = ef ‘a(x)dx ____ ef—l dx = 8—1.
Multiply both sides of our equation by the integrating
factor and note that the left-hand side of the resulting equation is the derivative of a product. e_‘y' — e'xy 2 22:6" (e‘xyy = 2xe" Integration by parts yields
foe" dx = 2.766" — f 26x = 2xe‘ — ge‘ + C. Consequently, e“)! = 2256" — 29" + C
y(x) = 2xe2" — 2.22" + cex The initial condition provides
3 = y(0) = 2(0)e2(°) — 2e2<0> + Ceo = —2 + C. Consequently, C = 5 and y(x) 2 he” —2e2x+56". ...

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