lecture_13 (dragged) 2

# lecture_13 (dragged) 2 - MA 36600 LECTURE NOTES MONDAY...

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MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 16 3 If the characteristic equation a r 2 + b r + c = 0 has two distinct real roots r 1 and r 2 , then the general solution to the di ff erential equation is y ( t ) = c 1 e r 1 t + c 2 e r 2 t for some constants c 1 and c 2 . Recall that a quadratic equation a r 2 + b r + c = 0 has two distinct real roots if its discriminant is positive: b 2 4 a c > 0. Example. Say that we wish to find the general solution to the di ff erential equation y + 5 y + 6 y = 0 . To solve this problem, we consider the characteristic equation. In this case, it is r 2 + 5 r + 6 = 0 . The polynomial factors as r 2 + 5 r + 6 = ( r + 2) ( r + 3) = r 1 = 2 , r 2 = 3 . Since these two roots are distinct, the general solution to the di ff erential equation is y ( t ) = c 1 e 2 t + c 2 e 3 t for some constants c 1 and c 2 . Proof of the General Solution. We return to the claim made earlier about the general solution to the homogeneous linear second order di ff erential equation with constant coe cients. In fact, we make a stronger claim. Consider the initial value problem
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