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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!

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  • $\begingroup$ 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. $\endgroup$ Jul 23, 2019 at 18:37
  • $\begingroup$ 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. $\endgroup$
    – Aphelli
    Jul 23, 2019 at 21:48
  • $\begingroup$ @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... $\endgroup$
    – dalta
    Jul 26, 2019 at 7:22

1 Answer 1

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Simple helpful facts:

  • 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.

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  • $\begingroup$ 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? $\endgroup$
    – dalta
    Jul 27, 2019 at 16:47
  • $\begingroup$ 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.. $\endgroup$ Jul 27, 2019 at 16:58

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