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

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Unformatted text preview: MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 16 3 If the characteristic equation a r2 + b r + c = 0 has two distinct real roots r1 and r2 , then the general solution to the differential equation is y (t) = c1 er1 t + c2 er2 t for some constants c1 and c2 . Recall that a quadratic equation a r2 + b r + c = 0 has two distinct real roots if its discriminant is positive: b2 − 4 a c > 0. Example. Say that we wish to find the general solution to the differential equation y ￿￿ + 5 y ￿ + 6 y = 0. To solve this problem, we consider the characteristic equation. In this case, it is r2 + 5 r + 6 = 0. The polynomial factors as r2 + 5 r + 6 = (r + 2) (r + 3) =⇒ r1 = −2, r2 = −3. Since these two roots are distinct, the general solution to the differential equation is y (t) = c1 e−2t + c2 e−3t for some constants c1 and c2 . Proof of the General Solution. We return to the claim made earlier about the general solution to the homogeneous linear second order differential equation with constant coefficients. In fact, we make a stronger claim. Consider the initial value problem a y ￿￿ + b y ￿ + c y = 0; where y (t0 ) = y0 , ￿ y ￿ (t0 ) = y0 . If the characteristic equation a r2 + b r + c = 0 has two distinct real roots r1 and r2 , then the solution to the differential equation is the function y (t) = c1 er1 t + c2 er2 t in terms of the constants c1 = ￿ y0 − r2 y0 −r1 t0 e r1 − r 2 and c2 = ￿ y0 − r1 y0 −r2 t0 e . r2 − r1 We sketch why this result is true. The proof relies on the “existence” and “uniqueness” of the solution to an initial value problem. For now, we give existence i.e., we simply explain why the function above is a solution to the differential equation. We have the derivatives y = c1 er1 t + c2 er2 t y￿ = r1 c1 er1 t + r2 c2 er2 t y ￿￿ = r1 2 c1 er1 t + r2 2 c2 er2 t This gives the expression ￿ ￿ ￿ ￿ ￿ ￿ a y ￿￿ + b y ￿ + c y = a r1 2 c1 er1 t + r2 2 c2 er2 t + b r1 c1 er1 t + r2 c2 er2 t + c c1 er1 t + c2 er2 t ￿ ￿￿ ￿ = a r1 2 c1 er1 t + b r1 c1 er1 t + c c1 er1 t + a r2 2 c2 er2 t + b r2 c2 er2 t + c c2 er2 t ￿ ￿ ￿ ￿ = a r1 2 + b r1 + c c1 er1 t + a r2 2 + b r2 + c c2 er2 t . Since r1 and r2 are roots of the characteristic equation, we have ￿ a r1 2 + b r1 + c = 0 =⇒ a y ￿￿ + b y ￿ + c y = 0. a r2 2 + b r2 + c = 0 Hence y is a solution to the differential equation as claimed. It suffices then to explain how to find the constants c1 and c2 from the initial conditions. We have the derivatives ￿ ￿ y (t) = c1 er1 t + c2 er2 t y0 = c1 er1 t0 + c2 er2 t0 =⇒ ￿ y ￿ (t) = r1 c1 er1 t + r2 c2 er2 t y0 = r1 c1 er1 t0 + r2 c2 er2 t0 ...
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