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Lecture Notes in Differential Equations - Bruce E. Shapiro

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42 LESSON 5. BERNOULLI EQUATIONS<br />

Multiply<strong>in</strong>g by dt and <strong>in</strong>tegrat<strong>in</strong>g,<br />

∫ ( d<br />

ze 2t2) ∫<br />

dt =<br />

dt<br />

4te 2t2 dt (5.26)<br />

Hence<br />

ze 2t2 = e 2t2 + C (5.27)<br />

From the <strong>in</strong>itial condition z(0) = 16,<br />

16e 0 = e 0 + C =⇒ 16 = 1 + C =⇒ C = 15 (5.28)<br />

Thus<br />

z = 1 + 15e −2t2 (5.29)<br />

From equation 5.16, y = z 1/4 , hence<br />

y =<br />

(1 + 15e −2t2) 1/4<br />

(5.30)

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