Lycée Gustave Eiffel Équations différentielles

sin(2x−3)
sin(x) cos(x)
cos2(3x)
tan(x)
2x−1
x2−x−1
1
xln(x)
ln(x)
xx
1 + x
1
1 + x
xpx2+ 1
2x+ 3
(x2+ 3x−4)2
2x+ 1
(x2+x+ 1)3
x
√x2+ 2
x
(2x2+ 3)n
n∈N
xx2
x2sin(x3)
sin(x)cos(x)
1
x
x2
1
1+4x2
1
1+2x2
1
4 + x2
1
2 + x2
1
√x2+ 5
1
√4x2−1
ln(x)
x2 2x
(x2+ 2x+
2) cos(x)
xcos(x)
a b c 1
x3−x=a x+b x−1+c x+ 1
1
x3−x
1
(1 + x2)n
n∈NFn
1
(1 + x2)n
F1
x2
(1 + x2)2
F2
n∈NFn+1 Fn
F3F4
y0+y= 0
y0+ (2x+ 1)y= 0
y0+1
√x2+ 1y= 0
y0+1
√x2−1y= 0
y0−(x)y= 0
y0+xy =x
y0+ 2xy −x−x2= 0
2y0−y+x= 0
(1 + x2)y0+xy =x+ 1
(x)y0+ (x)y=1
1 + x2
(1 + x2)y0+ 4xy = 0
xy0−y+ ln x= 0
x2y0+y−1=0
(1 −x2)y0+ (1 + x2)y=x
(x+ 1)2(xy0−y)+2x+ 1 = 0
xny0−αy = 0 (n∈N∗, α ∈R∗
+)
(E)y0−y=x√y
z=√y
y(E)z
(E0)
(E0)
(E)
(E)f(x+y) = xf(y) + yf(x)
fR
f(E)λ∈Rf
(Eλ)f0−f=λx
f(0) = 0
(Eλ)
(E)

f0(x)f(−x) = 1 (E)f
R
f(E)
g(x) = f(x)f(−x)
y0=Cy
C6= 0
(E)
y00 + 3y0+ 2y= 2x2+ 1
y00 + 4y0−5y=2x
y00 + 5y0−6y=x
y00 −2y0+ 2y=x
y00 −4y0+ 3y= 2 x
y00 +y= cos(2x)
y00 +y0−6y=xx
y00 −4y0+ 3y= 2 x
y00 + 2y0+y= (x2+ 1) −x
y00 −2y0+ 5y=xx
y00 + 4y0+ 13y=−x
y00 −2y0−8y=x4x
y00 −y0−2y= cos(x) + 3 sin(x)
y00 −y= cos(x)
y00 −4y0+ 5y=xsin(x)
y00 +y= cos(2x)
y00 −2y0+ 2y= cos2(x)
y00 −2y0+y=xxcos(2x)
y00 −3y0+ 2y=x+1+ x
y00 −2y0+ 2y= (x) cos(x)
y00 −y0−2y= (6x−1) 2x+ 4x2x
y00 −2y0+ 2y= (2 + 4x)xsin(x)
4y00 + 4y0+y= (x3+ 1) −x
2
y00 +y= sin(ωx)ω∈R
y00 −2ay0+ (a2+ 1)y= sin(x) + xax
a∈R
x0(t) = y(t)
y0(t) = x(t)
x0(t) = x(t)+4y(t)
y0(t) = x(t) + y(t)
x0(t)=5x(t)−6y(t) + t
y0(t)=4x(t)−5y(t) + t
x00 =x0+y0−y
y00 =x0+y0−x
(E)y00 + 6y0+ 9y=
−3x
√1 + x2
R(E0)
z(x) = 3xy(x) (E0)
(E)
(E0) (E)
y00 −2y0+y=xln(x)y00 + 4y0+ 4y=
x−2xln(x)y00 −4y0+4y=2x2x
1 + x2
(E)x2y00 + 3xy0+ 5y= 0
z(t) = y(t)y(E) ]0; +∞[
z
(E) ]0,+∞[
(E) ] − ∞,0[
(E)R
R+∗R−∗
x2y00 + 4xy0+ 2y= 0 x2y00 +xy0+y= 0 x2y00 + 3xy0+y= 0
(k, λ)∈R∗×R
Rx
f0(x) = k f(λ−x)
f00 (x)−f(−x) = sin(x) (E)
fR
fR
g h
g(x) = f(x) + f(−x)h(x) = f(x)−f(−x)
f0f00 g h
f(E)
g h
y00 −y= 0 (E1)
y00 +y= 2 sin(x) (E2)
(E1) (E2)
λ µ
g h λ µ
f λ µ
(E)