By Lerner M. E., Repin O. A.
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Extra resources for A boundary value problem for mixed-type equations in domains with multiply connected hyperbolicity subdomains
G. that t dy d2 y + y = (t2 − 1) exp − (t + 1) dt2 dt t2 2 = 0. com 61 Calculus 4c-4 Euler’s differential equation 2) a) Guessing. When we count the degrees, we see that if y is a polynomial of degree n, then the left hand side of (12) is again a polynomial of degree n. Hence we guess on y = at2 + bt + c, where d2 y dy = 2at + b and 2 = 2a. dt dt Then by insertion into the left hand side of the equation, t dy d2 y −(t+1) +y 2 dt dt = 2at−(t+1)(2at+b)+at2 +bt+c = −at2 +(2a−b−2a+b)t+c−b = −at2 + c − b.
The corresponding characteristic equation, λ2 − 3λ + 2 = (λ − 1)(λ − 2) = 0, has the solutions λ1 = 1 and λ2 = 2, so we conclude that the homogeneous equation has the complete solution x = c1 eu + c2 e2u = c1 t + c2 t2 , c1 , c2 ∈ R; t ∈ R+ . Then we can either ﬁnd a particular solution by guessing or by using the Wro´ nskian method. a) Guessing. The right hand side 2e2u is a solution of the homogeneous equation. We therefore guess on x = c · ue2u instead. Then dx = 2cue2u +ce2u , du x = cue2u , d2 x = 4cue2u +4ce2u , dt2 which give by insertion dx d2 x + 2x = 4cue2u +4ce2u −6cue2u −3ce2u +2cue2u = ce2u .
Com 44 Calculus 4c-4 Linear differential equations of second order with variable coefficients and thus d2 y − y = eu , du2 u = ln t, t > 0. The characteristic equation R2 − 1 = 0 has the roots R = ±1, so the homogeneous equation has the complete solution 1 y = c1 eu + c2 e−u = c1 t + c2 , t t > 0, where c1 and c2 are arbitrary constants. 2) Since the right hand side is of the form eu = ϕ1 (u), we guess a solution of the form y = aueu where dy = a(u + 1)eu du og d2 y = a(u + 2)eu . du2 Then by insertion, d2 y − y = a(u + 2)eu − aueu = 2aeu , du2 1 which is equal to eu for a = , and the complete solution is 2 1 1 1 y = ueu + c1 eu + c2 e−u = t ln t + c1 t + c2 , t > 0, 2 2 t where c1 and c2 are arbitrary constants.