Solution of Exam on Ordinary Differential Equations. Trial Aleph ...

4.2 Answer
y (x) =
1
p
2 (e
2x
+ e x )
(29)
5 Problem 4
y 00 2y 0 + 2y = e x cos x (30)
5.1 <strong>Solution</strong>
5.1.1 <strong>Solution</strong> <strong>of</strong> homogeneous equation
y 00 2y 0 + 2y = 0 (31)
, <strong>Solution</strong> is: 1 i; 1 + i
2 2 + 2 = 0 (32)
5.1.2 First way(Conjecture Method)
y = e x (C 1 cos x + C 2 sin x) (33)
e x cos x = Re e (1+i)x ; = 1 + i is a root <strong>of</strong> the characteristic equation (32) <strong>of</strong>
multiplicity 1, so the partial solution <strong>of</strong> equation (30) is obtained by conjecture
y p (x) = xe x (A cos x + B sin x) (34)
y 0 p (x) = d
dx (xex (A cos x + B sin x)) =
= A (cos x) e x + B (sin x) e x + Ax (cos x) e x + Bx (cos x) e x Ax (sin x) e x +
Bx (sin x) e x =
= Ae x cos x + Be x sin x + (A + B) xe x cos x + ( A + B) x sin x
y 0 p (x) = e x (A cos x + B sin x + (A + B) x cos x + ( A + B) x sin x) (35)
y 00
p (x) = d2
dx 2 (xe x (A cos x + B sin x)) =
= d
dx (ex (A cos x + B sin x + (A + B) x cos x + ( A + B) x sin x)) =
= 2A (cos x) e x +2B (cos x) e x 2A (sin x) e x +2B (sin x) e x +2Bx (cos x) e x
2Ax (sin x) e x =
= (2A + 2B) e x cos x + (2B 2A) e x sin x + 2Bxe x cos x 2Axe x sin x
y 00
p (x) = 2e x ((A + B) cos x + (B A) sin x + Bx cos x Ax sin x) (36)
y 00
p 2y 0 p + 2y p =
= 2e x ((A + B) cos x + (B A) sin x + Bx cos x Ax sin x)
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