Complex Variables and Partial Differential Equations (3130005)

Question-5a

BE | Semester-3 Winter-2019 | 26-11-2019

Q5) (a)

Using method of separation of variables, solve $\frac{\partial u}{\partial x} = 2 \frac{\partial u}{\partial t} + u$ .

Let,

u (x, y) = X (x) \cdot Y (y) = X \cdot Y

\Rightarrow \frac{\partial u}{\partial x} = u_{x} = X^{'} \cdot T & \frac{\partial u}{\partial t} = u_{t} = X \cdot T^{'}

Now,

\frac{\partial u}{\partial x} = 2 \frac{\partial u}{\partial t} + u

\Rightarrow X^{'} \cdot T = 2 X \cdot T^{'} + X \cdot T

→ Dividing by

X \cdot T

both the sides,

\Rightarrow \frac{X^{'} \cdot T}{X \cdot T} = 2 \frac{X \cdot T^{'}}{X \cdot T} + \frac{X \cdot T}{X \cdot T}

\Rightarrow \frac{X^{'}}{X} = 2 \frac{T^{'}}{T} + 1 = K

∎ \frac{X^{'}}{X} = K \Rightarrow \frac{1}{X} dX = K dx \Rightarrow \log X = Kx + C \Rightarrow X = c_{1} e^{Kx}

∎ 2 \frac{T^{'}}{T} + 1 = K \Rightarrow \frac{1}{T} dT = (\frac{K - 1}{2}) dt \Rightarrow \log T = (\frac{K - 1}{2}) t + C \Rightarrow T = c_{2} e^{(\frac{K - 1}{2}) t}

Solution:

u (x, y) = X \cdot T = (c_{1} e^{Kx}) (c_{2} e^{(\frac{K - 1}{2}) t})

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