Question

Non homo undetermined coefficients

Answer 1

@SithsAndGiggles

Answer 2

yp is asinx + bcosx, right?

Answer 3

Homogeneous solutions: Auxiliary equation:
\[r^3-r^2+r-1=r^2(r-1)+(r-1)=(r^2+1)(r-1)=0~~\implies~~r=\pm i,1\]
The fact that \(r=\pm i\) gives a solution of the form \(C_1\cos x+C_2\sin x\), we'll have to make an adjustment to what you suggested for the trial solution. Namely, \(y_p=x(A\sin x+B\cos x)\). You have derivatives
\[\begin{cases}\begin{align*}
y'&=x(A\cos x-B\sin x)+(A\sin x+B\cos x)\\
&=(Ax+B)\cos x+(-Bx+A)\sin x\\\\
y''&=A\cos x-(Ax+B)\sin x-B\sin x+(-Bx+A)\cos x\\
&=(-Ax+2B)\sin x+(-Bx+2A)\cos x\\\\
y'''&=-A\sin x+(-Ax+2B)\cos x-B\cos x-(-Bx+2A)\sin x\\
&=(-Ax+B)\cos x+(Bx-3A)\sin x\end{align*}
\end{cases}\]
There's a good chance I might have made a silly algebraic mistake...

Answer 4

yes then why is it that the particular soln of the correct answer is c1e^2x?

Answer 5

That definitely can't be right, unless you're missing some coefficients on one of the \(D\)s.

Answer 6

If the given solution is taken from a textbook, there can be a few things wrong with it. Could be a typo, which happens often, as a result of the author making a mistake in his/her work. It could be the answer to a different ODE altogether. Who knows? In any case, the given answer is not correct.

Answer 7

yes you're right. so ill just substitute the derivatives to the original equation

Answer 8

my yp is axe^-x + bx + c

Answer 9

am i right?

Answer 10

\[(D^3+D^2-4D-4)y=0\]
gives aux. eq.
\[r^3+r^2-4r-4=r^2(r+1)-4(r+1)=(r-2)(r+2)(r+1)=0\]
with roots \(r=\pm 2,-1\), so the homogeneous solution is
\[C_1e^{2x}+C_2e^{-2x}+C_3e^{-x}\]
Since the nonhomogeneous portion contains \(e^{-x}\), you need to modify the trial solution to contain an independent solution, \(xe^{-x}\), which you've done, so yes you're right.

Answer 11

thank you so much

Answer 12

Hm. Can you help me again? im having my exam tomorrow. ill post another question. Is it okay?

Answer 13

The homogeneous portion is fairly simple as it has a typical Euler-Cauchy form. Setting \(y=x^r\) gives
\[x^3(r(r-1)(r-2)x^{r-3})+2x^2(r(r-1))x^{r-2}-rxx^{r-1}+x^r=0\]
Dividing out \(x^r\) gives
\[\begin{align*}r(r-1)(r-2)+2r(r-1)-r+1&=0\\\\
r^3-3r^2+2r+2r^2-2r-r+1&=0\\\\
r^3-r^2-r+1&=0\\\\
r^2(r-1)-(r-1)&=0\\\\
(r^2-1)(r-1)&=0\\\\
(r+1)(r-1)^2&=0
\end{align*}\]
with roots \(r=-1,1_{(2)}\) (where \({}_{(2)}\) denotes the multiplicity). This means you have homogeneous solution \(y_c=\dfrac{C_1}{x}+C_2x+C_3x\ln x\).

Answer 14

I recently came across this pdf, which should help decide a trial solution for the \(3x^2\ln x\) on the RHS: http://www.maa.org/sites/default/files/DeLeon2010.pdf

According to the link, we should use \(y_p=(A+B\ln x)x^2\), which has derivatives
\[\begin{cases}\begin{align*}
Dy_p&=(2A+B)x+2Bx\ln x\\
D^2y_p&=2A+3B+2B\ln x\\
D^3y_p&=\dfrac{2B}{x}
\end{align*}\end{cases}\]
Subbing into the ODE gives
\[\begin{align*}
3x^2\ln x&=2Bx^2+2(2A+3B)x^2+4x^2B\ln x\\
&\quad-(2A+B)x^2-2Bx^2\ln x+(A+B\ln x)x^2\\\\
&=(3A+7B)x^2+3Bx^2\ln x
\end{align*}\]
which tells us that \(B=1\) and so \(A=-\dfrac{7}{3}\).

Answer 15

For RHS \(=-8x\), we can use the trial solution \(y_p=Ax\ln^2x\), with derivatives
\[\begin{cases}\begin{align*}
Dy_p&=A\ln^2x+2A\ln x\\
D^2y_p&=\dfrac{2A}{x}(\ln x+1)\\
D^3y_p&=-\dfrac{2A}{x^2}\ln x
\end{align*}\end{cases}\]
Subbing into ODE:
\[\begin{align*}
-8x&=-2Ax\ln x+4Ax(\ln x+1)-Ax\ln^2x-2Ax\ln x+Ax\ln^2x\\\\
&=4Ax
\end{align*}\]
which tells us that \(A=-2\).

Answer 16

Thank you @SithsAndGiggles soso much