Question

The general solution of $\frac{\partial u}{\partial x} = 4\frac{\partial u}{\partial y}$, given $u(0,y) = 8e^{-3y}$ is

• A. \( u(x,y) = 3e^{-3(4x+y)} \)
• B. \( u(x,y) = 8e^{12x-3y} \)
• C. \( u(x,y) = -3e^{-12x+3y} \)
• D. \( u(x,y) = 8e^{-12x-3y} \)

Hint:

For a linear first-order PDE of the form $a \frac{\partial u}{\partial x} + b \frac{\partial u}{\partial y} = 0$, the general solution is $u(x,y) = f(bx - ay)$, where $f$ is an arbitrary function. In this problem, $a=1$ and $b=-4$ (from $\frac{\partial u}{\partial x} - 4\frac{\partial u}{\partial y} = 0$), so the solution is $u(x,y) = f(-4x - 1y) = f(-4x-y)$ or equivalently $f(4x+y)$. Then apply the initial condition. This shortcut is based on the method of characteristics.

Answer 1

The Correct Option is D

The given partial differential equation (PDE) is a first-order linear PDE: $$ \frac{\partial u}{\partial x} = 4\frac{\partial u}{\partial y} $$ This can be rewritten as: $$ \frac{\partial u}{\partial x} - 4\frac{\partial u}{\partial y} = 0 $$ This is a homogeneous linear PDE of the form $P \frac{\partial u}{\partial x} + Q \frac{\partial u}{\partial y} = R$, where $P=1$, $Q=-4$, and $R=0$. We can solve this using the method of characteristics. The characteristic equations are: $$ \frac{dx}{P} = \frac{dy}{Q} = \frac{du}{R} $$ $$ \frac{dx}{1} = \frac{dy}{-4} = \frac{du}{0} $$ 
Step 1: Find the first characteristic curve
From $\frac{dx}{1} = \frac{dy}{-4}$: $$ -4 \, dx = dy $$ Integrate both sides: $$ \int -4 \, dx = \int dy $$ $$ -4x = y + c_1 $$ $$ y + 4x = c_1 $$ Let $\xi = y+4x$. 
Step 2: Find the second characteristic curve (relation for $u$)
From $\frac{du}{0}$, it implies that $du = 0$, which means $u$ is a constant along the characteristic curves. So, $u$ is a function of the first characteristic constant $c_1$: $$ u(x,y) = f(y+4x) $$ where $f$ is an arbitrary function. 
Step 3: Use the initial condition to find the specific function $f$
Given the initial condition $u(0,y) = 8e^{-3y}$. Substitute $x=0$ into the general solution: $$ u(0,y) = f(y+4(0)) = f(y) $$ We are given $u(0,y) = 8e^{-3y}$. So, $f(y) = 8e^{-3y}$. 
Step 4: Substitute $f$ back into the general solution
Replace $y$ in $f(y)$ with $(y+4x)$: $$ u(x,y) = 8e^{-3(y+4x)} $$ $$ u(x,y) = 8e^{-3y-12x} $$ $$ u(x,y) = 8e^{-12x-3y} $$