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Lecture Notes in Differential Equations - Bruce E. Shapiro

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196 LESSON 22. NON-HOMOG. EQS.W/ CONST. COEFF.<br />

and the particular solution, <strong>in</strong> either case, is<br />

∫ (∫<br />

)<br />

y P = e r2t e (r1−r2)t q(t)e −r1t dt dt (22.90)<br />

Return<strong>in</strong>g to the orig<strong>in</strong>al ODE (22.67), before we def<strong>in</strong>ed q(t) = f(t)/a,<br />

y P = 1 ∫ (∫<br />

)<br />

a er2t e (r1−r2)t f(t)e −r1t dt dt (22.91)<br />

We have just proven the follow<strong>in</strong>g two theorems.<br />

Theorem 22.6. The general solution of the second order homogeneous<br />

l<strong>in</strong>ear differential equation with constant coefficients,<br />

ay ′′ + by ′ + cy = 0 (22.92)<br />

is<br />

y H =<br />

{<br />

C 1 e r1t + C 2 e r2t if r 1 ≠ r 2<br />

(C 1 t + C 2 )e rt if r 1 = r 2 = r<br />

where r 1 and r 2 are the roots of the characteristic equation<br />

(22.93)<br />

ar 2 + br + c = 0 (22.94)<br />

Theorem 22.7. A particular solution for the second order l<strong>in</strong>ear differential<br />

equation with constant coefficients<br />

ay ′′ + by ′ + cy = f(t) (22.95)<br />

is<br />

y P = 1 a er2t ∫<br />

e (r1−r2)t (∫<br />

)<br />

f(t)e −r1t dt dt (22.96)<br />

where r 1 and r 2 are the roots of the characteristic equation<br />

ar 2 + br + c = 0 (22.97)<br />

Example 22.2. Solve the <strong>in</strong>itial value problem<br />

⎫<br />

y ′′ − 9y = e 2t ⎪⎬<br />

y(0) = 1<br />

⎪⎭<br />

y ′ (0) = 2<br />

(22.98)

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