# Question about possible solutions to a differential equation

It is my second week studying differential equations, and I am struggling with the following question:

Which of the following statements are correct?

1. There is a second order homogenous linear differential equation with continuous coefficients in the open interval $$(\frac{π}{2},\frac{3π}{2})$$, of which the functions $$x$$ and $$sinx$$ are solutions.

2. There is a second order homogenous linear differential equation with continuous coefficients in the open interval $$(\frac{π}{2},\frac{3π}{2})$$, of which the functions $$x$$ and $$x^2$$ and $$x(x+2)$$ are solutions.

3. There is a second order homogenous linear differential equation with continuous coefficients in the open interval $$(\frac{-π}{2},\frac{π}{2})$$, of which the function $$x^2$$ is a solution.

4. There is a second order homogenous linear differential equation with constant coefficients in the open interval $$(\frac{π}{2},\frac{3π}{2})$$, of which the functions $$x$$ and $$x^2$$ and $$x(x+2)$$ are solutions.

5. There is a second order non-homogenous linear differential equation with continuous coefficients in the open interval $$(0,2)$$, to which the functions $$x$$ and $$x^2$$ and $$x^3$$ are solutions.

I am pretty sure that 1. and 4. are incorrect, but am struggling with the rest. Could anybody please advise?

Thank you!

• Can you give a reason for your observations? Look at $y(0)$ and $y'(0)$ for $y(x)=x^2$ and $y(x)=x-\sin x$ and what solution the IVP with these values would have. Jul 23, 2019 at 18:37
• 1, 2. Try to write some. 3. Similar to 5, but evaluate in $0$. 4. There is a specific structure theorem for solutions of second-order homogenous ODEs with constant coefficients. 5. For solutions of such a (hypothetical) ODE, $y \longmapsto (y(1),y’(1))$ is an affine isomorphism. Jul 23, 2019 at 21:48
• @LutzL 1 looks wrong for me because the general solution to a second order homogenous equation is $c1e^ax+c2e^bx$. For 4., as far as I have understood, the laws of the Wronskian do not apply if the coefficients are constant. But again, I started studying this topic two weeks ago, so I am not sure about anything as of yet... Jul 26, 2019 at 7:22

• i) At a regular point (coefficients all continuous) of an explicit homogeneous second order ODE $$y''+py'+qy=0,$$ if both $$y(x_0)=0$$ and $$y'(x_0)=0$$, then $$y=0$$ on the whole interval of coefficient continuity.

• ii) Given any two functions $$y_1$$, $$y_2$$, you get a homogeneous second order ODE having them as solutions as $$0=\det\pmatrix{y_1&y_2&y\\y_1'&y_2'&y'\\y_1''&y_2''&y''}.$$ Singular points are the roots of the leading coefficient $$y_1y_2'-y_1'y_2$$.

• iii) Given any three functions $$y_1$$, $$y_2$$, $$y_3$$, you get a second order ODE having them as solutions as $$0=\det\pmatrix{1&1&1&1\\y_1&y_2&y_3&y\\y_1'&y_2'&y_3'&y'\\y_1''&y_2''&y_3''&y''}.$$ Singular points are again the roots of the leading coefficient $$\det\pmatrix{1&1&1\\y_1&y_2&y_3\\y_1'&y_2'&y_3'}.$$

• iv) If an explicit homogeneous linear ODE has constant coefficients, then it is completely determined by the roots of its characteristic polynomial. If a solution contains a term $$p(x)e^{\lambda x}$$, then $$λ$$ is a root of multiplicity at least $$1+\deg p$$.

1. By ii), the only singular points of the ODE for these functions are the roots of $$x\cos x-\sin x$$, of which there is one in the interval by the intermediate value theorem.

2. The third function is a linear combination of the first two, and the leading coefficient in ii) is $$2x^2-x^2=x^2$$, so that only $$x=0$$ has to be excluded from the domain.

3. See i)

4. By iv), $$0$$ needs to be characteristic root of multiplicity at least 3. Which is impossible.

5. By iii), such an ODE exists. The given interval can be a domain if the leading coefficient \begin{align} \det\pmatrix{1&1&1\\x&x^2&x^3\\1&2x&3x^2} &=x\det\pmatrix{1&1&1\\1&x&x^2\\1&2x&3x^2} =x\det\pmatrix{1&1&1\\0&x-1&x^2-1\\0&x&2x^2} \\[1em] &=x^2(x-1)\det\pmatrix{1&x+1\\1&2x} =x^2(x-1)^2 \end{align} has no roots in it, which is obviously not the case.

• Many thanks! I would have never thought of such an approach. I think I have understood the explanation as well. So this basically leaves only 2. as correct, right? Jul 27, 2019 at 16:47
• Yes. It seems strange that the non-trigonometric functions there are combined with multiples of $\pi$ as interval bounds, in 5. this was not the case, so it is not really a structural design decision for this task.. Jul 27, 2019 at 16:58