All of the solution of $y^{(33)}(x)-y^{(30)}(x)=e^{-x}$

Question

Find all of the solution of the following differential equation s.t. $\lim_{x\to +\infty}\frac{y(x)}{x^2}=0$ $$y^{(33)}(x)-y^{(30)}(x)=e^{-x}$$

I have immediately dismissed the idea of using the variation of parameters method, since I need 33 linearly independent solutions of the homogeneous; however, putting it into a system is leading nowhere. Any hint?

Answer 1

I would try this: $$y^{(33)}(x)-y^{(30)}(x)=e^{-x}$$ $$y^{(31)}(x)-y^{(28)}(x)=e^{-x}+C_1x+C_2$$

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$$y^{(5)}(x)-y^{(2)}(x)=e^{-x}+\sum_{i=0}^{27}C_{i}x^{i}$$ $$y^{(3)}(x)-y(x)=e^{-x}+\sum_{i=0}^{29}C_{i}x^{i}$$ The differential equation is of order 33 so you need 33 constants.

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Since the differential equation is linear you can solve this apart: $$y^{(3)}(x)-y(x)=e^{-x}$$ The characteristic polynomial is: $$\implies r^3-1=0 \implies (r-1)(r^2+r+1)=0$$ $$\implies r=1, r=-\dfrac 12 \pm i\dfrac {\sqrt 3}2$$ The solution of the homogeneous equation is: $$y(x)=e^{-x/2} \left (A \cos (\dfrac {\sqrt 3 x}2)+B\sin(\dfrac {\sqrt 3 x}2) \right)+Ce^x$$ For the particular solution try: $$y_p(x)=Fe^{-x}$$ $$-Fe^{-x}-Fe^{-x}=e^{-x} \implies F=-\frac 12$$ So the general solution of the DE is: $$y(x)=e^{-x/2} \left (A \cos (\dfrac {\sqrt 3 x}2)+B\sin(\dfrac {\sqrt 3 x}2) \right)+Ce^x-\frac 12e^{-x}$$

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In absence of initial conditions the solution is: $$y(x)=e^{-x/2} \left (A \cos (\dfrac {\sqrt 3 x}2)+B\sin(\dfrac {\sqrt 3 x}2) \right)+Ce^x-\frac 12e^{-x}+\sum_{n=0}^{29}a_{n}x^{n}$$

Answer 2

Hint: Do you know any functions which, if you differentiate them $30$ times, give you zero? That should dispose of most of your linearly independent solutions for the homogeneous equation.

[I think the question is a good one for testing understanding, so once you have solutions, you will have to think carefully]

Answer 3

If we let $$U=\frac{d^{30}y}{dx^{30}}$$

Then we can begin first by solving the following equation:

$$\frac{d^3 U}{dx^3}-U=e^{-x}$$

$$\left(\frac{d^3 }{dx^3}-1\right)U=e^{-x}$$

$$U=-\frac{1}{2}e^{-x}$$

It then must follow that

$$\frac{d^{30}y}{dx^{30}}=-\frac{1}{2}e^{-x}$$

$$y=-\frac{(-1)^{30}}{2}e^{-x}+\sum_{m=0}^{29}C_{m}x^{m}$$

$\left[ C_{m} \right]$ is a set of 30 arbitrary constants.

Answer 4

Hint In terms of $u := y^{(30)}$, the differential equation becomes $$\phantom{(\ast)} \qquad u''' - u = e^{-x} . \qquad (\ast)$$

Trying the ansatz $u = A e^{-x}$ gives a particular solution $u_0$.

The homogeneous equation corresponding to $(\ast)$ is $$u''' - u = 0 ,$$ which can be solved using the usual method of finding the roots of the characteristic equation $$r^3 - 1 = 0 ,$$ giving three linearly independent solutions $v_1, v_2, v_3$.

Thus, the general solution to $(\ast)$ is $$u_0 + A v_1 + B v_2 + C v_3 ,$$ and we need to solve $$y^{(30)} = u_0 + A v_1 + B v_2 + C v_3 .$$ But notice that $u_0'' = u_0$ and (rearranging the homogeneous equation above) $v_i''' = v_i$, so we can write $$y^{(30)} = u_0^{(30)} + A v_1^{(30)} + B v_2^{(30)} + C v_3^{(30)} = (u_0 + A v_1 + B v_2 + C v_3)^{(30)},$$ and integrating gives the general solution $$y = u_0 + A v_1 + B v_2 + C v_3 + \sum_{j = 0}^{29} a_j x^j$$ of the differential equation in $y$. (This solution has 33 parameters, as expected.) All that remains is to do is check which solutions satisfy the asymptotic condition $$\lim_{x \to \infty} y = 0 ,$$ but the forms of $u_0, v_1, v_2, v_3$ make this straightforward.