differensial ligning
Posted: 14/12-2011 22:52
[tex]\frac{ds}{dt}=600-\frac{2s}{200+t}[/tex]
[tex]\frac{ds}{dt}+\frac{2s}{200+t}=600[/tex]
[tex]\frac{ds}{dt}e^{ln(100+t/2)}+\frac{2s}{200+t}e^{ln(100+t/2)}=e^{ln(100+t/2)}600[/tex]
[tex]\frac{d}{dt}(se^{ln(100+t/2)})=(100+t/2)600[/tex]
[tex]se^{ln(100+t/2)}=\int(100+t/2)600dt[/tex]
[tex]s(100+t/2)=600(100t+t^2/4)+C[/tex]
t=0 s=20 000
[tex]20.000\cdot100=C[/tex]
Jeg prøvde å bruke produktregelen for derivasjon baklengs
fasiten oppgave 3:
http://wiki.math.ntnu.no/_media/tma4100 ... -16_lf.pdf
[tex]\frac{ds}{dt}+\frac{2s}{200+t}=600[/tex]
[tex]\frac{ds}{dt}e^{ln(100+t/2)}+\frac{2s}{200+t}e^{ln(100+t/2)}=e^{ln(100+t/2)}600[/tex]
[tex]\frac{d}{dt}(se^{ln(100+t/2)})=(100+t/2)600[/tex]
[tex]se^{ln(100+t/2)}=\int(100+t/2)600dt[/tex]
[tex]s(100+t/2)=600(100t+t^2/4)+C[/tex]
t=0 s=20 000
[tex]20.000\cdot100=C[/tex]
Jeg prøvde å bruke produktregelen for derivasjon baklengs
fasiten oppgave 3:
http://wiki.math.ntnu.no/_media/tma4100 ... -16_lf.pdf