Kernel and Differential Operator

Question

I hope your day is great so far. I have a question. I am given a Cauchy-Euler ODE $$\big((x+2)^3\mathbf{D}^3+4(x+2)^2\mathbf{D}^2+3(x+2)\mathbf{D}+1\big)y=\frac{1}{x+2},$$ where $\mathbf{D}^n=d^n/dx^n$. Letting $e^t=x+2$ transforms this ODE, such that the coefficients are constants, giving a new ODE $$(\mathsf{D}^3+\mathsf{D}^2+\mathsf{D}+1)y=e^{-t},$$ where $\mathsf{D}^n=d^n/dt^n$. I can't use the formula $$\frac{1}{p(\mathsf{D})}[e^{at}]=\frac{1}{p(a)}e^{at},p(a)\neq0,$$ since $p(a)=0$. How can I find the particular solution then? I actually find that we can use $$y_p=\frac{1}{p_{n-k}(a)}\bigg(\frac{1}{k!}t^k+\text{ker}\,\mathsf{D}^k\bigg)e^{at},$$ for this case. However, I don't understand this formula and was hoping someone could explain that. Thanks in advance!

Answer 1

$$(\mathsf{D}^3+\mathsf{D}^2+\mathsf{D}+1)y=e^{-t}$$ $$y=\dfrac 1 {(\mathsf{D}^3+\mathsf{D}^2+\mathsf{D}+1)}e^{-t}$$ $$y=\dfrac 1 {(D+1)(D^2+1)}e^{-t}$$ $$y=\dfrac 1 {(D+1)}\dfrac 1{(D^2+1)}e^{-t}$$ You can apply the first formula since we have $D^2+1=(-1)^2+1=2 \ne 0$: $$y=\dfrac 12\dfrac 1 {(D+1)}e^{-t}$$ $$y=\dfrac {e^{-t}}2\dfrac 1 {D}1$$ $\dfrac 1 D$ is simly an integration: $$y=\dfrac {te^{-t}}2$$

Answer 2

Your transformed equation $$(\mathsf{D}^3+\mathsf{D}^2+\mathsf{D}+1)y=e^{-t}$$ has $e^{-t}$ as solution to the homogeneous part of the ODE, so proper use of the Methos of undetermined coefficients is trying a solution $(At+B)e^{-t}$. Substituting in the equation you will find $$y_p=(\frac{1}{2}t+ \frac{1}{2})e^{-t}$$