Author Topic: FE2  (Read 574 times)

Victor Ivrii

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« on: December 08, 2014, 04:13:04 PM »
Find the general solution of
x^4 y^{(4)} + 6 x^3 y^{(3)} + 7 x^2 y^{(2)} + x y' - y = 3 \ln x + \cos (\ln x) \, .

It is Euler's equation. Its characteristic polynomial is
r(r-1)(r-2)(r-3)+6r(r-1)(r-2) +7r(r-1)+r -1=\\
r^4 - 6r^3 +11r^2 -6r + 6r^3 -18r^2 +12r +7r^2-7r+r-1=
with characteristic roots $r_{1,2}=\pm 1$, $r_{3,4}=\pm i$ and plugging $t=\ln x$ we arrive to
y^{(4)}_t-y= 3t + \cos (t).
Solution to homogeneous equation is
z= C_1 e^t +C_2 e^{-t}+C_3 \cos (t)+ C_4\sin (t)=\\
 C _1 x +C_2 x^{-1}+ C_3\cos (\ln x) + C_4\sin(\ln x)
and the particular solution to inhomogeneous equation is $y_p = y_{p1}+y_{p2}$ with $y_{p1}= at +b$ and $y_{p2}=  (c\cos (t) +d \sin(t))t$ solving equation with right hand expressions $f_1=3\ln x$ and $f_2= \cos(\ln x)$ respectively.

Plugging $y_{p1}$ we get
-at -b=3t \implies a=-3,b=0\implies y_{p1}=-3t=-3\ln x
and plugging $y_{p2}$ we get
3(c\sin (t)-d\cos (t)) = \cos(t) \implies c=0,d=-\frac{1}{3}\implies \\
y_{p2}=-\frac{1}{3}\sin (t) t=-\frac{1}{3}\sin (\ln x)\ln x.
Adding (\ref{eq-2-2})--(\ref{eq-2-4}) we get
y= C _1 x +C_2 x^{-1}+ C_3\cos (\ln x) + C_4\sin(\ln x)-3\ln x -\frac{1}{3}\sin (\ln x)\ln x.
« Last Edit: December 14, 2014, 10:35:47 PM by Victor Ivrii »