We must know that x=e^(t) and t=lnx. Professor says we cannot complete the problem any other way.
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We must know that x=e^(t) and t=lnx. Professor says we cannot complete the problem any other way.
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x^2y''-xy'+y=2x Must use cauchy-euler, and x=e^t and t=lnx, and also make it homogeneous by using annihilator operators. I have it good to annhiliator operators, but when I find a particular solution, I do not get the same resultant :/
Answer is : general solution: y=c1(x) + c2(xlnx) +c3 x(lnx)^2
Through the annihilator operator:
D(D-1)^(3) * y= 0
auxillary eqn:
m(m-1)^3=0
therefore, m1=m2=m3=1 and m4= 0
therefore,
y=c1(e^t) +c2t(e^t) + c3t^2(e^t) +c4 e^(0t) Note: This is without plugging in the fact that t=lnx. Therefore, I made a mistake somewhere between finding the annihilator operator and finding particular solution.
The problem is that your "annihilator" equation has constant coefficients and is NOT the "Cauchy-Euler" equation given. Also, did you notice that the original differential equation was second order while your "annihilator" equation was fourth order, and your "solution" has four independent functions?
Taking t= ln(x), as your teacher suggested, $\displaystyle \frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}$. And then $\displaystyle \frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)=$$\displaystyle -\frac{1}{x^2}\frac{dy}{dt}+$$\displaystyle \frac{1}{x}\left(\frac{1}{x}\frac{d^2y}{dt^2}\righ t)= $$\displaystyle \frac{1}{x^2}\frac{d^2y}{dt^2}- \frac{1}{x^2}\frac{dy}{dt}$.
So $\displaystyle x^2y''-xy'+y=2x$ becomes the "constant coefficients" equation $\displaystyle \frac{d^2y}{dt^2}- \frac{dy}{dt}- \frac{dy}{dt}+ y= \frac{d^2y}{dt^2}- 2\frac{dy}{dt}+ y= 2e^t$.
The corresponding characteristic equation is $\displaystyle r^2- 2r+ 1= (r- 1)^2= 0$ so the general solution to the corresponding homogeneous equation is $\displaystyle y_h(t)= C_1e^t+ C_2te^t$.
The right hand side is $\displaystyle e^t$ and, normally, we would try a specific solution of the entire equation of the form $\displaystyle y= Ae^t$ but that and $\displaystyle xe^t$ are already solution to the homogeneous equation so we try $\displaystyle y= At^2e^t$. Then $\displaystyle y'= 2Ate^t+ At^2e^t$ and $\displaystyle y''= 2Ae^t+ 4Ate^t+ At^2e^t$. Putting those into the differential equation $\displaystyle \frac{d^2y}{dt^2}- 2\frac{dy}{dt}+ y= 2e^t$, we have $\displaystyle 2Ae^t+ 4Ate^t+ At^2e^t- 4Ae^t- 2At^2e^t+ At^2e^t= 2Ae^t= 2e^t$ so $\displaystyle A= 1$.
The general solution to they entire equation is $\displaystyle y(t)= C_1e^t+ C_2te^t+ t^2e^t$.
Since we used the substitution $\displaystyle t= ln(x)$ that becomes $\displaystyle y(x)= C_1e^{ln(x)}+ C_2 ln(x)e^{ln(x)}+ (ln(x))^2e^{ln(x)}= C_1x+ C_2x ln(x)+ x (ln(x))^2$.
Annihilator on right side: would be (D-1)^2.
Annihilator on left side for : 2e^t is the form of (D-alpha)^n, where in this case, alpha is 1, and n=1, therefore, (D-1)^1,.... but we need to remember D kills 3, so right side: D(D-1)
So if we combine that together, then, D(D-1)^3 * y=0 is the new homogeneous equation. However, you are saying I am wrong because I am assuming e^t can have a constant coefficient?
I did notice that, that is why I felt my answer was awkward. So are you saying it is not possible to annihilate this equation?
I understand the latter, of finding yp, yp' and yp'', however I do not understand why you just make the assumption that y=At^2e^t. Are y_h(t)= c1e^t+c2te^t, but this is our complimentary function. We need to find our particular solution. The 2 ways I know how to find particular solutions are: use annihilator operator or set particular solution, yp=u1y1+u2y2, which is methods of variation of parameters. She says specifically that she wants x=e^t used and annihilator approach.
So we must be able to annihilate this right?
Taking a look from a section earlier I had an equation:
y''-2y'-3y=4e^(x) - 9
so for this problem:
associated homogenous:
r''-2r-3=0
(r-3)(r+1)=0
r1=3, and r2=-1
therefore, our complimentary solution is: y=c1e^3x+ c2e^-x
Now we annihilate the right side (Just taking auxiliary equation):
(D-3)(D+1)(r-3)(r+1)=0
Annihilate left side:
D(D-1)4e^x - 9=0
Therefore,
D(D-1)(D-3)(D+1) * y=0 (and now it is homogenous)
So then, our r3=1 and r4=0, which are our particular solution.
so y=c1e^3x + c2e^-x + c3e^x + c4e^(0)
Then solve for yp, yp' and get general solution.
Why cannot we do that for this equation? I do not understand.
Hi, I was right with regards to the annihilator operator, I found out what I did wrong. There is a c4, but when we find yp, yp', and yp'', we find out that that c4 goes away by finding out that it = 0.