2.6 KiB
#start of lecture 11
last lecture we did cauchy euler equations:
ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0
where a,\ b,\ c are still constants and \in \mathbb{R}
x=e^ta{\frac{d^2y}{dt^2}}+(b-a){\frac{dy}{dt}}+cy=f(e^t)<- lousy notation, the y here isnt quite the same as in the above definition.y=x^rar^2+(b-a)r+C=0three cases: (i)r_1\ne r_{2}then:y_{h}(x)=c_{1}x^{r_{1}}+c_{2}x^{r_{2}}(ii)r_{1}=r_{2}=rthen:y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)(iii)r_{1,2}=\alpha+i\betathen:y_{h}(x)=x^2(c_{1}\cos(\ln \beta x)+c_{2}\sin \ln(\beta x))now find one particular solution for a non homogenous soultion, using variation of parameters, combine the y_h and y_p to get y(x).
not all equations can fall into cauchy euler type.
y''+p(x)y'+q(x)y=f(x) (1) <- no general solution procudure always
but, if y_{1}(x) solves y''+p(x)y'+q(x)y=0
then we can find the general solution to the non homogenous equation (1) by guessing it in the form y(x)=v(x)y_{1}(x)
y'=v'y_{1}+vy_{1}'
y''=v''y_{1}+2v'y_{1}'+vy_{1}''
(v''y_{1}+2v_{1}'y_{1}'+y_{1}''v)+p(x)(v'y_{1}+vy_{1}')+q(x)vy_{1}=f(x)
v\cancelto{ 0 }{ (y_{1}''+p(x)y_{1}'+q(x)y_{1}) }+v''y_{1}+(2y_{1}'+p(x)y_{1})v'=f
y_{1}v''+()
v''+\left( \frac{2y_{1}'}{y_{1}}+p \right)v'=\frac{f}{y_{1}}
v'=u
u'+\left( \frac{2y_{1}'}{y_{1}}+p \right)u=\frac{f}{y_{1}}<- this is a linear first order equation
how to solve linear first order equation? we compute the integrating factor \mu
\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1})^2}e^{\int P(x) \, dx}=y_{1}^2e^{\int p(x) \, dx}
isnt this nice? some kind of magic. We made some guesses and we arrived somewhere.
#ex find the general solution to the equation:
y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}
if y_{1}(x)=e^{-x^2} is one solution.
therefore were finding the solution of the form: y(x)=v(x)y_{1}=v(x)e^{-x^2}
v'=u
u'+\left( \frac{2y_{1}'}{y_{1}}+4x \right)u=\frac{8{e^{-x^2}e^{-2x}}}{e^{-x^2}} <-(p(x)=4x)
u'+\left( \frac{2{e^{-x^2}(-2x)}}{e^{-x^2}}+4x \right)u=8e^{-2x}
u'=8e^{-2x}
u=-4e^{-2x}+c_{1}
v'=u=-4e^{-2x}+c_{1}
v(x)=2e^{-2x}+c_{1}x+c_{2}
general solution:
y(x)=v(x)y_{1}(x)=(2e^{-2x}+c_{1}x+c_{2})e^{-x^2}Free vibrations
mr^2+br+k=0 characteristic polynomail
(i) r_{1}\ne r_{2} b^2-4mk>0
y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}
r_{1,2}=-\frac{b}{2m}\pm \frac{\sqrt{ b^2-4mk }}{2m}<0
then the limit of the homogenous solution is 0 as t->\infty (over damped case)
(ii) r_{1}=r_{2}=-\frac{b}{2m}
r_{1}=r_{2}=-\frac{b}{2m}
y_{h}(t)=e^-\frac{b}{2m}+c_{2}te^{-b/2m}t limit =0 as t approches inf critically damped