‚é™—pitul—tif de l— sé—n™e n¦Q Équ—tions différentielles

y0+ 2y= 4e12xE
fRf(x) = 4xe12xE
f0+ 2f= 4e12x
f0(x)=4e12x+ 4x(2)e12x= 4e12x8xe12x= 4e12x2f(x)
f0(x) + 2f(x) = 4e12x2f(x) + 2f(x) = 4e12xf
E
(E0)y0+ 2y= 0 y0=2y
f0=af f(x) = Ceax CR
(E0)y(x) = Ce2xCR
gRE g f(E0)
g E g0+ 2g= 4e12x
(gf)0+ 2(gf)=(g0+ 2g)(f0+ 2f) = 4e12x4e12x= 0
gf(E0)
gf(E0)gRE
gf(E0) (gf)0+ 2(gf) = 0
(gf)0+ 2(gf) = (g0+ 2g)(f0+ 2f)=(g0+ 2g)4e12x= 0 (g0+ 2g)=4e12x
g E
g E
g E g
gf(E0) (E0)
y(x) = Ce2xCRgf=Ce2xE
g(x) = Ce2x+f(x) = Ce2x+ 4xe12x=e2x(C+ 4xe)CR
E C
2e0
g0(0) = 2e g0(0) = e2·0(C+ 4 ·0·e) = C C =2e
g0(x) = e2x(2e+ 4xe)=2e2x+1(1+2x)
(E)y00 2y0+y=x2
f(x) = x2+ 4x+ 6 (E)
fR
f0(x) = 2x+ 4
f00(x) = 2
f00(x)2f0(x) + f(x)=24x8 + x2+ 4x+ 6 = x2
f(x) = x2+ 4x+ 6 (E)
2Rp(x) = c+bx+ax2
(a, b, c)R3p(E)p00 2p0+p=x2
p0(x) = 2ax +b
p00(x) = 2a
p00 2p0+p=x22a4ax 2b+c+bx +ax2= (2a2b+c)+(b4a)x+ax2=x2
ax2=x2a= 1
x(b4a) = 0 (b4·1) = 0 b= 4
(2a2b+c) = 0 (2 ·12·4 + c)=0 c= 6
2 (E)f(x) = x2+4x+6
g(x) = (2x5)ex+x2+ 4x+ 6 (E)
g(x) = r(x) + f(x)r(x) = (2x5)ex
(r+f)00 2(r+f)0+ (r+f)=(r00 2r0+r)+(f00 2f0+f)=(r00 2r0+r) + x2
g(E)
(r+f)00 2(r+f)0+ (r+f) = x2
r00 2r0+r= 0 E0
r(x) = (2x5)ex
r0(x) = 2ex+r(x)
r00(x) = 2ex+r0(x)
r00 2r0+r= 2ex+r0(x)2r0(x) + r(x)=2exr0(x) + r(x) = 2ex2exr(x) + r(x)=0
r E0g(E)
f0=af a R
f(x) = Ceax CR
f0(x0) = Ceax0=y0C C =y0
eax0
C f0C
f0(x) = y0
eax0·eax
2f0=f2f(1) = 3
f0=1
2f f(x) = Ce 1
2x
CR
2f(1) = 3 f(1) = 3
2f(1) = Ce1
2Ce1
2=3
2
C=3
2e1
2
=3e1
2
2f0(x) =
3e1
2
2e1
2x=3
2e1
2(x+1) =3
2e(x+1)
i(Eq)
Li0(t) + Ri(t) = E
uf0+vf =w(u, v, w)Rf0=af +b
(a, b)R
f0=af +b(a, b)R
f(x) = Ceax b
a
CR
Li0(t) + Ri(t) = E f 0=af +b
i0(t) = R
Li(t)+ E
Li(t) = CeR
Lt
E
L
R
L
=CeR
Lt+E
R
CRt= 0 i(0) = CeR
L·0+E
R= 0
C=E
Ri(t) = E
ReR
Lt+E
R=E
ReR
Lt+ 1
i(t)t+
R
L<0 lim
t+R
Lt=−∞
lim
t+R
Lt=−∞ lim
t+
eR
Lt= 0
lim
t+
eR
Lt= 0 lim
t+
eR
Lt+ 1 = 1
lim
t+
eR
Lt+ 1 = 1 lim
t+
E
ReR
Lt+ 1=E
R
i(t)t+E
R
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