Discussion
Consider the following differential equation
\(\frac{\partial y}{\partial x} = 3\frac{\partial y}{\partial t} + y\)
If \(y(x, 0) = 10e^{-2x}\), then the solution of the differential equation is
Answer: A
The solution is \(y(x,t) = 10e^{-2x-t}\).
Try \(y = 10e^{-2x + kt}\), which matches the initial condition at \(t = 0\). Substituting:
\[ \begin{aligned} -2y &= 3ky + y \\ 3k &= -3,\quad k = -1 \end{aligned} \]
So \(y = 10e^{-2x-t}\). Check: \(\partial y/\partial x = -2y\), and \(3\,\partial y/\partial t + y = -3y + y = -2y\). With \(k = +1\) the right side is \(4y\), and with \(k = \pm 2\) it is \(-5y\) or \(7y\); none equals \(-2y\).
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