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Use the Frobenius method to solve $x^2y''+xy'+y=0$ about $x=0$.

$x=0$ is a singular regular point. My problem is the roots of the indicial equation are $-i$ and $i$. What do I do with complex roots???

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  • $\begingroup$ This is a Cauchy-Euler equation, so it has a solution of the form $y=x^r$. Substituting this and its derivatives for $y,y',y''$ gives you an auxiliary equation, also with roots $r=\pm i$. Solutions of this form would look like $$y=C_1\cos(\ln x)+C_2\sin(\ln x)$$. $\endgroup$ – user170231 Apr 1 '15 at 17:13
  • $\begingroup$ I need to solve it using the frobenius method though. $\endgroup$ – guest10923 Apr 1 '15 at 17:22
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So you have the roots $\pm i$, normally you order them by largest real part, but obviously the real parts are the same, so lets set $\alpha_1=i,\alpha_2=-i$. Then the first solution is $x^i\sum_{k=0}^\infty a_kx^k$ where $x^i=e^{i\ln x}=\cos(\ln(x))+i\sin(\ln(x))$, so $$y_1=[\cos(\ln(x))+i\sin(\ln(x))]\sum_{k=0}^\infty a_kx^k$$ substituting into the ODE (for simplicity of calculations let $y_1=\sum_{k=0}^\infty a_kx^{k+i}$) we get $$x^2y_1''+xy_1'+y_1=\sum_{k=0}^\infty[(k+i)(k+i-1)+(k+i)+1]a_kx^{k+i}=0$$ so $$[k^2+2ki-1-i+k+i-1]a_k=0$$ i.e. $$[k^2+(2i+1)k]a_k=0,\,\forall k=0,1,2,...$$ since we can set $a_0=1$ WLOG, for all other indices we must have $a_k=0$, thus we get $y_1=x^i=\cos(\ln(x))+i\sin(\ln(x))$

For the second solution, notice that $\alpha_1-\alpha_2=2i$ which is not an integer, so $y_2=x^{-i}\sum_{k=0}^\infty b_kx^k$, and similarly we obtain $y_2=\cos(\ln(x))-i\sin(\ln(x))$

Now we can write any solution as some linear combination of $y_1,y_2$, i.e. $y=A\cos(\ln(x))+B\sin(\ln(x))$

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