Question

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

• A. \( y(x,t) = 10e^{-2x-t} \)
• B. \( y(x,t) = 10e^{-2x+t} \)
• C. \( y(x,t) = 10e^{-2x-2t} \)
• D. \( y(x,t) = 10e^{-2x+2t} \)

Hint:

Try a solution of the form y = 10 e^{-2x+ct} and find the constant c that satisfies the equation.

Answer 1

The Correct Option is A

Instead of solving the equation from scratch, we can plug each candidate answer into $\frac{\partial y}{\partial x} = 3\frac{\partial y}{\partial t} + y$ and keep only the one that actually satisfies it along with $y(x,0) = 10e^{-2x}$.

1. $y = 10e^{-2x-t}$: $y_x = -2y$, $y_t = -y$. Check: $3y_t+y = -3y+y = -2y = y_x$. This works, and at $t=0$ it gives $10e^{-2x}$, so it fits the initial data too.
2. $y = 10e^{-2x+t}$: $y_x=-2y$, $y_t=y$. Then $3y_t+y = 3y+y=4y$, but $y_x=-2y$, so $-2y \ne 4y$ unless $y=0$. Fails the PDE.
3. $y = 10e^{-2x-2t}$: $y_x=-2y$, $y_t=-2y$. Then $3y_t+y=-6y+y=-5y \ne -2y$. Fails.
4. $y = 10e^{-2x+2t}$: $y_x=-2y$, $y_t=2y$. Then $3y_t+y=6y+y=7y \ne -2y$. Fails.

Only the first form balances the equation on both sides, so option A, $y(x,t) = 10e^{-2x-t}$, is the correct solution.