What is the the minimum possible order of a homogeneous linear ODE with real constants having $x^2 \sin(x) $ as a solution?
Solution:It is given as if $x^2 \sin(x)$ is a solution then $\iota $ is root of the ODE which is repeated three times,but as complex roots occur in conjuagte pairs so $-\iota$ is also a root with multiplicity $ \ 3$.Hence,the minimum order should be 6.
I'm not getting how $x^2 \sin(x)$ is a solution implies $\iota $ is root with multiplicity 3?
Please explain this
Consider the differential equation $$ y''(x) - 2y'(x) + y(x) = 0. \tag{1} $$ The associated characteristic equation is $\lambda^2 - 2\lambda + 1 = (\lambda - 1)^2$ and its root is $\lambda = 1$, repeated twice. Thus the general solution of $(1)$ is $$ y(x) = Ae^{x} + Bxe^{x} $$ where the additional $"x"$ on the second term is due to $\lambda = 1$ being a repeated root of multiplicity 2. A similar argument as above shows that the general solution of $$ y'''(x) - 3y''(x) + 3y'(x) + y(x) = 0 $$ is $$ y(x) = Ae^{x} + Bxe^{x} + Cx^2e^{x}, $$ where the additional $"x"$ on the second term and $"x^2"$ on the third term is due to $\lambda = 1$ being a repeated root of multiplicity 3 in this case.
Now, for the general solution to involve $\sin(x)$, the characteristic equation must have complex root $\lambda = i$. Since a solution is given as $x^2\sin(x)$, this means that $\lambda = i$ must repeated three times. This would indicate that the minimum order is 3, except this is false since complex root appears in conjugate pairs, i.e. if $\lambda = a + bi$ is a root of the associated characteristic equation, then $\lambda = a - bi$ must also be a root. Consequently, the minimum order of a homogeneous linear ODE with real constants is $3\times 2 = 6$.