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

Step 1: Try an exponential solution.
The equation \( \dfrac{\partial y}{\partial x} = 3\dfrac{\partial y}{\partial t} + y \) is linear, so try \( y(x,t) = 10e^{-2x}g(t) \), matching the given shape at \( t=0 \) where \( g(0)=1 \).
Since the initial condition is \( y(x,0)=10e^{-2x} \), it makes sense to guess the \( t \) dependence is also a simple exponential, \( g(t) = e^{ct} \) for some constant \( c \).

Step 2: Substitute the guess \( y = 10e^{-2x+ct} \) into the PDE.
\( \dfrac{\partial y}{\partial x} = -2y \) and \( \dfrac{\partial y}{\partial t} = cy \).
Putting these into the equation gives \( -2y = 3(cy) + y \).

Step 3: Solve for \( c \).
Divide through by \( y \) (nonzero): \( -2 = 3c+1 \), so \( 3c = -3 \) and \( c = -1 \).

Step 4: Write the full solution.
With \( c=-1 \), \( y(x,t) = 10e^{-2x-t} \), and this reduces to \( 10e^{-2x} \) when \( t=0 \), matching the given condition.

Final Answer:
This exponential form solves both the PDE and the initial condition. \[ \boxed{y(x,t) = 10e^{-2x-t}} \]