2nd order linear ODE
Posted: 10/05-2014 01:07
gitt:
[tex](1-x)y''\,+\,xy'\,-\,y=1-x[/tex]
[tex]y_h\,\,[/tex] går greit å løse:
[tex](1-x)y''\,-\,(1-x)y'\,+\,y'\,-\,y=0[/tex]
[tex](1-x)(y''\,-\,y')\,=y\,-\,y'[/tex]
[tex]\int\frac{y''-y'}{y'-y}\,dy=\int\frac{dx}{x-1}[/tex]
[tex]\ln|y' - y|=\ln|x-1|+C'[/tex]
[tex]y' - y=C*(x-1)[/tex]
integrerende faktor etc gir så:
[tex]y_h=Cx\,+\,De^x\,+\,E[/tex]
hvordan bestemmes så [tex]\,\,y_p\,\,[/tex]?
[tex](1-x)y''\,+\,xy'\,-\,y=1-x[/tex]
[tex]y_h\,\,[/tex] går greit å løse:
[tex](1-x)y''\,-\,(1-x)y'\,+\,y'\,-\,y=0[/tex]
[tex](1-x)(y''\,-\,y')\,=y\,-\,y'[/tex]
[tex]\int\frac{y''-y'}{y'-y}\,dy=\int\frac{dx}{x-1}[/tex]
[tex]\ln|y' - y|=\ln|x-1|+C'[/tex]
[tex]y' - y=C*(x-1)[/tex]
integrerende faktor etc gir så:
[tex]y_h=Cx\,+\,De^x\,+\,E[/tex]
hvordan bestemmes så [tex]\,\,y_p\,\,[/tex]?