# Download e-book for kindle: Calculus 1c-5, Examples of Simple Differential Equations II by Mejlbro L.

By Mejlbro L.

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Extra resources for Calculus 1c-5, Examples of Simple Differential Equations II

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I. The characteristic polynomial is R2 + aR + b. 1) If s is not a root of the characteristic polynomial, we guess on c · est . Then by insertion d2 x dx + bx = c(s2 + as + b)est . +a dt2 st The assumption assures that s2 + as + b = 0, hence this is equal to est for c = a particular solution is in this case given by s2 1 , and s2 + as + b 1 est . com 33 Calculus Analyse 1c-5 Linear differential equations og second order and of constant coefficients 2) If s is a root of the characteristic polynomial, we of course get zero by insertion.

Guess a complex exponential x = c · e(1+2i)t . Although it is not required, we shall nevertheless also solve the homogeneous equation. I. If we put x = c · e(1+2i)t into the left hand side of the equation, we get d3 x d2 x dx +x + 3 +3 3 2 dt dt dt = c (1 + 2i)3 + 3(1 + 2i)2 + 3(1 + 2i) + 1 e(1+2i)t = c{1 + (1 + 2i)}3 e(1+2i)t = 8c(1 + i)3 e1+2i)t = 8c · 2i(1 + i)e(1+2i)t = 16(−1 + i)c e(1+2i)t . This is equal to e(1+2i)t for c = equation is Re −1 − i (1+2i)t e 32 =− 1 −1 − i 1 · = , so a solution of the corresponding real 16 −1 + i 32 1 t 1 t 1 t e Re (1 + i)e2it = e sin 2t − e cos 2t.

We see that R2 + 1 is a divisor. Then we get the factorization R6 + 9R4 + 24R2 + 16 = (R2 + 1)(R4 + 8R2 + 16) = (R2 + 1)(R2 + 4)2 . Thus the complex roots are ±i (simple roots) and ±2i (double roots). The complete solution is x = c1 cos t + c2 sin t + c3 cos 2t + c4 sin 2t + c5 t cos 2t + c6 t sin 2t, where c1 , c2 , c3 , c4 , c5 , c6 ∈ R are arbitrary constants, and t ∈ R. 4 Find a particular solution of the diﬀerential equation d3 x d2 x dx + x = et cos 2t, + 3 +3 dt3 dt2 dt t ∈ R. A. Linear inhomogeneous diﬀerential equation of third order and of constant coeﬃcients.