Finding the solution for an operator in an equation

Question

Problem

Find the fundamental solution about the operator $$z=\frac{d3}{dx^3}-7 \frac{d}{dx}+6$$ and check if it's true

Solution

$y'''=-7y'(x)+6y(x)=0$

$y(0)=0$

$y'(0)=0$

$y''(0)=1$

The characteristic equation:

$\lambda^{3}-7 \lambda + 6 =0$

$\lambda = \{1,2,-3\}$

Fundamental system of solutions: $ C_{1}e^{x}+C_{2}e^{2x}+C_{3}e^{-3x}$

$y(x)=C_{1}e^{x}+C_{2}e^{2x}+C_{3}e^{-3x}$

$y'(x)=C_{1}e^{x}+2C_{2}e^{2x}-3C_{3}e^{-3x}$

$y''(x)=C_{1}e^{x}+4C_{2}e^{2x}+9C_{3}e^{-3x}$

Now for x=0

$y(0)=C_{1}+C_{2}+C_{3}=0$

$y'(0)=C_{1}+2C_{2}-3C_{3}=0$

$y''(0)=C_{1}+4C_{2}+9C_{3}=1$

When we solve the system we get: $C=\{-\frac{1}{4}, \frac{1}{5}, \frac{1}{20}\}$

$y(x)=-\frac{1}{4}e^{x}+\frac{1}{5}e^{2x}+\frac{1}{20}e^{-3x}$

$\varepsilon(x)=H(x)[-\frac{1}{4}e^{x}+\frac{1}{5}e^{2x}+\frac{1}{20}e^{-3x}]$

Now to check:

$L(\frac{d}{dx}\varepsilon)=\mu$ in $D'(R)$ and $<L\varepsilon,\varphi>=\varphi(0)$

$<L(\frac{d}{dx}) \varepsilon, \varphi>$ for $\varphi \in C^{\infty}_{0}(R)$

$<L(\frac{d}{dx}) \varepsilon, \varphi>= <\sum y(x)H(x),a_{j}(x) \varphi(x)>+<z^{(n)}(x)H(x)+\delta(x),a_{0}\varphi >$

$\sum_{j=0}^{n} <H(x),a_{j}(x)z^{(n-j)}(x) \varphi(x)>+<\delta,\varphi>=<H(x),(\sum a_j(x)z^{(n-j)}(x))\varphi > + < \delta, \varphi >$

$\varepsilon^{(k)}=y^{(k)}(x)H(x),k=\overline{0,n-1}$

$\varepsilon^{(k)}=y^{(k)}(x)H(x)+\delta(x) \in D'(R)$

$y^{(n-j)}(x)\in C^{\infty}_{0}(R),j=\overline{0,n}$

$H(x)$ is linear

If $\sum_{j=0}^{n} a_{j}(x)z^{(n-j)}(x)=0=><L(\frac{d}{dx})\varepsilon,\varphi>=<\delta,\varphi>$

$L(\frac{d}{dx})\varepsilon=\mu \in D'(R)$

$y(x)=C_{1}e^{x}+C_{2}e^{2x}+C_{3}e^{-3x}$

$y'(x)=C_{1}e^{x}+2C_{2}e^{2x}-3C_{3}e^{-3x}$

$y''(x)=C_{1}e^{x}+4C_{2}e^{2x}+9C_{3}e^{-3x}$

$y'''(x)=C_{1}e^{x}+8C_{2}e^{2x}-27C_{3}e^{-3x}$

Question:

Is my solution correct?

Answer 1

You can easily verify by using Kernels Decompostion Lemma :

You associated polynom is :

$$X^3-7X+6=(X-1)(X^2+X-6)=(X-1)(X-2)(X+3)$$

So since the primality , you have your solution space:

$$ \mathcal{S}=Ker(D-I_d)\bigoplus Ker(D-2I_d) \bigoplus Ker(D+3I_d)$$

with $D$ is the derivation operator

So it is correct.