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Lecture Notes in Differential Equations - Bruce E. Shapiro

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144 LESSON 17. SOME SPECIAL SUBSTITUTIONS<br />

Replac<strong>in</strong>g z with its orig<strong>in</strong>al value of z = y ′ ,<br />

Hence<br />

dy<br />

dt = Ce−6t (17.6)<br />

∫ dy<br />

y = dt (17.7)<br />

dt<br />

∫<br />

= Ce −6t dt = (17.8)<br />

= − 1 6 Ce−6t + C ′ (17.9)<br />

S<strong>in</strong>ce −C/6 is still a constant we can rename it C ′′ = −C/6, then rename<br />

C ′′ back to C, giv<strong>in</strong>g us<br />

Example 17.2. F<strong>in</strong>d the general solution o<br />

y = Ce −6t + C ′ (17.10)<br />

y ′′ + 6y ′ = t (17.11)<br />

We already have solved the homogeneous problem y ′′ + 6y ′ = 0 <strong>in</strong> example<br />

(17.1). From that we expect that<br />

where<br />

y = Y P + Y H (17.12)<br />

Y H = Ce −6t + C ′ (17.13)<br />

To see what we get if we substitute z = y ′ <strong>in</strong>to (17.11) and obta<strong>in</strong> the first<br />

order l<strong>in</strong>ear equation<br />

z ′ + 6z = t (17.14)<br />

An <strong>in</strong>tegrat<strong>in</strong>g factor is µ = exp (∫ 6dt ) = e 6t , so that<br />

Integrat<strong>in</strong>g,<br />

∫<br />

d (<br />

ze<br />

6t ) = (z ′ + 6t) e 6t = te 6t (17.15)<br />

dt<br />

d (<br />

ze<br />

6t ) ∫<br />

dt = te 6t dt (17.16)<br />

dt<br />

ze 6t = 1<br />

36 (6t − 1)e6t + C (17.17)<br />

z = 1 6 t − 1 36 + Ce−6t (17.18)

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