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I have stumbled upon this problem in my differential equations homework:

Does there exist a second order linear homogeneous differential equation of the form $$y''+ p(x)y' + q(x)y= 0$$ such that $ p(x) $ and $ q(x) $ are continuous on the entire real line that has both $ f(x) = \cos(x) $ and $ g(x) = e^{x^2} $ as solutions?

I have the general solution of $ y = c_1\cos(x) + c_2e^x $ and the $ y' $ and $ y'' $ of this function. I am guessing I need to combine the $ y, y', $ and $ y'' $ but I am having trouble canceling out the resulting constants. Any suggested paths or methods?

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    $\begingroup$ You know the cross product, right? $\endgroup$ – ccorn Mar 12 '17 at 7:46
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    $\begingroup$ How does that help in this context? @ccorn $\endgroup$ – Bryan Chen Mar 12 '17 at 8:09
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Set $y_1=f$, $y_2=g$. You want $$\begin{bmatrix}q(x)\\p(x)\\1\end{bmatrix}^\top \begin{bmatrix}y_i(x)\\y_i'(x)\\y_i''(x)\end{bmatrix} = 0 \qquad\text{for}\ i\in\{1,2\}$$ So you look for a vector $[q(x),p(x),1]$ whose dot product with two given vectors is zero. Well, that is what the cross product is for: Setting $$\begin{bmatrix}w(x)\\v(x)\\u(x)\end{bmatrix} = \begin{bmatrix}y_1(x)\\y_1'(x)\\y_1''(x)\end{bmatrix}\times \begin{bmatrix}y_2(x)\\y_2'(x)\\y_2''(x)\end{bmatrix}$$ Gives you functions $w,v,u$ such that $$u(x)\,y_i''(x) + v(x)\,y_i'(x) + w(x)\,y_i(x) = 0 \qquad\text{for}\ i\in\{1,2\}$$ and dividing by $u(x)$ should give you $$\begin{align} p(x) &= \frac{v(x)}{u(x)} & q(x) &= \frac{w(x)}{u(x)} \end{align}$$ The catch is that $u(x)$ may be zero for some $x$ and that $p$ or $q$ might have some non-removable singularities there. That would violate the requirement that $p,q$ be continuous on the entire real line.

Indeed, in the given case, there are infinitely many $x_k$ such that $u(x_k)=0$ (left as an exercise). That implies $$\begin{bmatrix}y_1(x_k)\\y_1'(x_k)\end{bmatrix} = \lambda_k\begin{bmatrix}y_2(x_k)\\y_2'(x_k)\end{bmatrix} \qquad \lambda_k = \frac{y_1(x_k)}{y_2(x_k)}$$ (using the fact that $y_2(x)=g(x)=\exp(x^2)$ is nonzero everywhere) and therefore, if the ODE were to hold at $x_k$ with continuous $p$ and $q$, we would have $$\begin{gather} p(x_k)\,y_1'(x_k) + q(x_k)\,y_1(x_k) = \lambda_k\left(p(x_k)\,y_2'(x_k) + q(x_k)\,y_2(x_k)\right) \\\therefore\quad y_1''(x_k) = \lambda_k y_2''(x_k) \end{gather}$$ and altogether $$\begin{bmatrix}y_1(x_k)\\y_1'(x_k)\\y_1''(x_k)\end{bmatrix} = \lambda_k\begin{bmatrix}y_2(x_k)\\y_2'(x_k)\\y_2''(x_k)\end{bmatrix}$$ But then the cross product would be a triple of zeros. In other words, whenever $u(x_k) = 0$, you also need $v(x_k)=0$ and $w(x_k)=0$, otherwise the ODE cannot be fulfilled with continuous $p$ and $q$ at those $x_k$. I leave it to you to check those conditions. Hint: Checking $x=0$ is easy.

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Simply put $f(x)$ and $g(x)$ into the equation to get $$-\cos x-p\sin x+q\cos x=0\\ (2+4x^2)e^{x^2}+2xe^{x^2}p+e^{x^2}q=0$$ Therefore $$\frac{q-1}p=\tan x\\ 2+4x^2+2xp+q=0$$ And you can find $p,q$ by solving these two nonlinear equations.

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  • $\begingroup$ oh wow, I see, thank you @jeckerya $\endgroup$ – Bryan Chen Mar 12 '17 at 7:48

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