Machine-translated by Google. Questions are authored in English.
Top

Discussion

Consider the differential equation \(x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}-4y=0\) with the boundary conditions of \(y(0)=0\) and \(y(1)=1\). The complete solution of the differential equation is

  • A.\(x^{2}\)
  • B.\(\sin\left(\frac{\pi x}{2}\right)\)
  • C.\(e^{x}\sin\left(\frac{\pi x}{2}\right)\)
  • D.\(e^{-x}\sin\left(\frac{\pi x}{2}\right)\)

Answer: A

The solution is \(y = x^{2}\).

This is a Cauchy-Euler equation. Try \(y = x^{m}\):

\[ \begin{aligned} m(m-1)+m-4 &= 0 \\ m^{2} &= 4,\ m = \pm 2 \\ y &= c_{1}x^{2}+c_{2}x^{-2} \end{aligned} \]

The condition y(0) = 0 needs \(c_{2}=0\), since \(x^{-2}\) is unbounded at the origin, and y(1) = 1 then gives \(c_{1}=1\).

The three sine options satisfy both boundary values but not the equation: substituting \(\sin(\pi x/2)\), for example, leaves terms in \(x^{2}\sin\) and \(x\cos\) that do not cancel.

No comment is present. Be the first to comment.
Loading…

Post your comment

Please log in or register to post a comment.