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Lecture Notes in Differential Equations - Bruce E. Shapiro

Lecture Notes in Differential Equations - Bruce E. Shapiro

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142 LESSON 16. LINEAR EQS. W/ CONST. COEFFICENTS<br />

Example 16.5. Solve the <strong>in</strong>itial value problem<br />

⎫<br />

y ′′ − 6y ′ + 9y = 0 ⎪⎬<br />

y(0) = 4<br />

⎪⎭<br />

y ′ (0) = 17<br />

(16.71)<br />

The characteristic equation is<br />

0 = r 2 − 6r + 9 = (r − 3) 2 (16.72)<br />

S<strong>in</strong>ce there is a repeated root r = 3, the general solution of the homogeneous<br />

equation is<br />

y = Ae 3t + Bte 3t (16.73)<br />

By the first <strong>in</strong>itial condition, we have 4 = y(0) = A.Hence<br />

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

From the second <strong>in</strong>itial condition,<br />

y = 4e 3t + Bte 3t (16.74)<br />

y ′ = 12e 3t + Be et + 3Bte 3t (16.75)<br />

17 = y ′ (0) = 12 + B + 3B = 12 + B =⇒ B = 5 (16.76)<br />

Hence the solution of the <strong>in</strong>itial value problem is<br />

y = (4 + 5t)e 3t . (16.77)

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