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Find the general solution of $y^{(6)}+2y^{(4)}+y'' = 0$.

$r^6+2r^4+r^2=0$

$r^2(r^4+2r^2+1)=0$

$r^2[(r^2+1)(r^2+1)]$

So we have the roots:

$0$: Multiplicity 2

$+i$: Multiplicity 2

$-i$: Multiplicity 2

Now I'm not sure. I'm supposed to arrive at:

$y(x) = c_1+c_2x+c_3\cos x+c_4\sin x+c_5x\cos x+c_6x\sin x$

EDIT: I'm particularly curious as to what the multiplicity does to the general solution.

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It's the multiplicity that necessitates the factor of $x$ in the 2nd, 5th, and 6th terms in the answer. –  Gerry Myerson Oct 6 '12 at 7:54
    
I'm confused by the "sixth degree" in the title and the $y^6$; this suggests a horrendous non-linear differential equation. Or should I read "sixth order" and $y^{(6)}$ (as well as $y^{(4)}$)? –  Marc van Leeuwen Oct 6 '12 at 12:12
    
Corrected to sixth order. –  GEdgar Oct 6 '12 at 12:29

3 Answers 3

$$y^{6}+2y^{4}+y^{2}=0$$ Using differential operator i.e. $\displaystyle \frac{dy}{dx}=Dy$, we get, $$ D^{6}y+2D^{4}y+D^{2}y=0.$$ $$\Rightarrow (D^{6}+2D^{4}+D^{2})y=0.$$ Therefore the auxiliary equation is, $$ D^{2}(D^{4}+2D^{2}+1)=0$$ $$\Rightarrow D^{2}(D^{2}+1)^{2}=0$$ $$\Rightarrow D=0,0,\pm i, \pm i$$ Hence the general solution of the given equation is, $$y(x)= C_{1}+C_{2}x+(C_{3}+C_{4}x)\cos x+(C_{5}+C_{6}x)\sin x.$$

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For the case $r=0$, you will get $y_1(x)=1$ and $y_2(x)=1$, but always remember that we need the solutions to be linearly independent. So, we assume $y_2(x)=u(x)y_1(x)$ and we substitute back in the differential equation to find $u(x)$. If you do that you will find $u(x)=x$. That implies $y_2(x)=x$. You do the same with the other cases of multiplicity. To form the general solution, you need all of the sixth solutions you have just found and write down the general solution as

$$ y(x)=c_1y_1(x)+c_2 y_2(x)+\dots+c_6y_6(x)\,. $$

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For every complex number $r$ and integer $n$, let $e_{r,n}:x\mapsto x^n\mathrm e^{rx}$. If $r$ is a root of multiplicity $k$ of the characteristic equation, then the function $e_{r,n}$ is a solution for every $0\leqslant n\leqslant k-1$. The family of functions $(e_{r,n})_{r,n}$ is linearly independent hence this yields a vector space of solutions whose dimension is the degree of the equation. That is, the exact and complete set of solutions.

Let $c_{t,n}:x\mapsto x^n\cos(tx)$ and $s_{t,n}:x\mapsto x^n\sin(tx)$. The formula in your post describes the general element of the vector space generated by $e_{0,0}$, $e_{0,1}$, $e_{\mathrm i,0}$, $e_{\mathrm i,1}$, $e_{-\mathrm i,0}$ and $e_{-\mathrm i,1}$ since, for every real number $t\ne0$, $e_{\mathrm it,n}$ and $e_{-\mathrm it,n}$ generate the same vector space as $c_{t,n}$ and $s_{t,n}$.

Finally, there exists some coefficients $\lambda_n$, $\mu_n$ and $\nu_n$, for $n=0$ and $n=1$, such that $$ y=\lambda_0e_{0,0}+\lambda_1e_{0,1}+\mu_0c_{1,0}+\mu_1c_{1,1}+\nu_0s_{1,0}+\nu_1s_{1,1}. $$

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