Exercices corrigés de mathématiques pour Licence de Biologie

e
2s+ 1
(s−2)(s2+ 1)
y00(x)−5
2y0(x) + y(x) = −5 2sin x , y(0) = 0, y0(0) = 2 .
y00 + (z0−y0) = −3
4y
z00 −(z0−y0) = −3
4z ,
y(0) = z(0) = 0 y0(0) = 1 z0(0) = −1
y0
1(x) + 2y0 2(x) + 3y0 3(x) = 0
y0
1(x)−y0 2(x) = 3x−3
y0
2(x) + 2y3(x) = 1 −x2,
y1(0) = y2(0) = y3(0) = 0
ω∈Rf(x) = xsin(ωx)
f00(x)=2ωcos(ωx)−ω2f(x)
f
s
(s2+ 3)2

2s+ 1
(s−2)(s2+ 1) =1 s−2−s s2+ 1
s2Y−2−5
2sY +Y=−5 2
1
s2+ 1 , Y =2s+ 1 (s−2)(s2+ 1) ,
y(x) = 2x−cos x
s2Y−1 + s(Z−Y) = −3
4Y
s2Z+ 1 −s(Z−Y) = −3
4Z , s2(Y+Z) = −3
4(Y+Z)
s2(Y−Z)−2+2s(Z−Y) = −3
4(Y−Z),
Y+Z= 0
2s2Y−2−4sY =−3
2Y , Y=1
(s−1/2)(s−3/2) =−1 s−1/2+1 s−3/2=−Z .
y(x) = −x/2+3x/2=−z(x)
sY1+ 2sY2+ 3sY3= 0
sY1−sY2= 3/s2−3/s
sY2+ 2Y3= 1/s −2/s3,
sY1=−2sY2−3sY3
−sY2−sY3= 1/s2−1/s
sY2+ 2Y3= 1/s −2/s3,
sY1=−2sY2−3sY3
−sY2−sY3= 1/s2−1/s
(2 −s)Y3= 1/s2−2/s3,
Y1=−2/s2+ 3s3
Y2= 1/s2
Y3=−1/s3.
y1(x) = −2x+3
2x2y2(x) = x y3(x) = −1 2x2
f0(x) = sin(ωx) + xω cos(ωx)
f00(x) = ωcos(ωx) + ωcos(ωx)−xω2sin(ωx) = 2ωcos(ωx)−ω2f(x).
f00
s2L(f)−sf(0) −f0(0) = 2ωs
s2+ω2−ω2L(f)L(f) = 2ωs (s2+ω2)2.
s
(s2+ 3)2=1
2√3
2√3s
s2+√322=Lx
2√3sin(√3x).