3.10) a)
y′−3x=1
y′=3x+1
y(x)=∫(3x+1)dx
y(x)=23x2+x+C
c)
y′−4x+x2=1
y′=−x2+4x+1
y=−3x3+2x2+x+C
e)
y′′+6x−3=0
y′′=−6x+3
y′=3x2+3x+C1
y=−x3+23x+C1x+C2
3.11)
b)
y=x+x2+C
y′=−2x+1
3.12)
b)
y′+x2=x+1;y(6)=0
c)
y′′−x+1=0;y(1)=0,y′(1)=0
umstellen
y′′=x−1
integrieren
y′2x2−x+C1
nochamal
y=6x3−2x2+C1∗x+C2
y′(1)=0:0=212−1+C1→C1=21
cC2=−61
spez lösung:
y=−6x3+2x2+2x−61
3.13)
f(1)=0
y′x1
y=ln∣x∣+C
0=ln∣1∣+C
y=ln∣x∣
3.15)
mittlere Beschl:
a(t)=0.5s2m
a(t)=v′(t)=s′′(t)
a, s(t)=?
a(t)=v′(t)=0,5
v(t)=0.5t+C1
algemeine lösung:
s(t)=4t2+C1t+C2
b)
v(0)=0
s(0)=0
0=40+C1∗0+C2→C2=0
0=0,5∗0+C1→C1=0
s(t)=4t2=0.25t2
c)
bis 200km/h = 55,5m/s
v(t)=55.5=0.5(t)→t=111.1s
3.38 c)
21∗p0=p0∗e−7.99h