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c ′ 1 (x) cos x + c′ 2 (x) sin x = 0,
−c ′ 1 (x) sin x + c′ 2 (x) cos x = 1/ sin x,
cos x
⇐⇒
− sin x
sin x c ′
1 (x)
cos x c ′ 2 (x)
=
0
1/ sin x
c ′ 1 (x) = −1 c1 = −x c ′ 2 (x) = cos x/ sin x
c2(x) = log | sin x| ¯y(x) = log | sin x| sin x − x cos x
y(x) = c1 cos x + c2 sin x − x cos x + log | sin x| sin x.
y ′′ +3y ′ +2y = 0 λ 2 +3λ+2 = 0
λ = −1 λ = −2 Φ(x, c1, c2) = c1e −x + c2e −2x
¯y(x) = c1(x)e −x + c2(x)e −2x
¯y ′ (x) = c ′ 1(x)e −x + c ′ 2(x)e −2x − c1(x)e −x − 2c2(x)e −2x .
c ′ 1 (x)e−x + c ′ 2 (x)e−2x = 0 ¯y ′ (x) = −c1(x)e −x − 2c2(x)e −2x
¯y ′′ (x) = −c ′ 1(x)e −x − 2c ′ 2(x)e −2x + c1(x)e −x + 4c2(x)e −2x .
−c ′ 1(x)e −x −2c ′ 2(x)e −2x +c1(x)e −x +4c2(x)e −2x −3c1(x)e −x −6c2(x)e −2x +c1(x)e −x +c2(x)e −2x = √ 1 + e x
−c ′ 1 (x)e−x − 2c ′ 2 (x)e−2x = √ 1 + e x
c ′ 1 (x)e−x + c ′ 2 (x)e−2x = 0,
−c ′ 1 (x)e−x − 2c ′ 2 (x)e−2x = √ 1 + e x ,
c ′ 1 (x) = ex√ 1 + e x c ′ 2 (x) = −e2x√ 1 + e x
c1(x) = e x√ 1 + ex
√1
dx = + t dt = z 1/2 dz = 2
3 z3/2 = 2
3 (1 + t)3/2 = 2
c2(x) = − e 2x√ 1 + ex
dx = − t √ 1 + t dt = − 2
3 t(1 + t)3/2 + 2
(1 + t)
3
3/2 dt
= − 2
3 t(1 + t)3/2 + 4
15 (1 + t)5/2 = − 2
3 ex (1 + e x ) 3/2 + 4
15 (1 + ex ) 5/2
¯y(x) = 4
15 (1 + ex ) 5/2 e −2x ,
y(x) = c1e −x + c2e −2x + 4
15 (1 + ex ) 5/2 e −2x .
3 (1 + ex ) 3/2
y ′′′ − 3y ′′ + 3y ′ − y = 0 λ 3 − 3λ 2 +
3λ − 1 = 0 (λ − 1) 3 = 0
Φ(x, c1, c2, c3) = c1ex + c2xex + c3x2ex
⎛
⎞ ⎛
⎠ ⎝ c′ 1 (x)
c ′ 2 (x)
c ′ 3 (x)
⎞ ⎛
⎠ = ⎝ 0
0
ex /x
⎝ ex xe x x 2 e x
e x e x (x + 1) xe x (2 + x)
e x e x (2 + x) e x (2 + 4x + x 2 )
A
det(A) = e 3x ((x+1)(x 2 +4x+2)+x 2 (x+2)+x 2 (x+2)−x 2 (x+1)−x(x+2) 2 −x(x 2 +4x+2)) = 2e 3x .
⎞
⎠