Complex Variables and Partial Differential Equations (3130005)

Question-5b

BE | Semester-3 Summer-2020 | 27-10-2020

Q5) (b)

Solve the p.d.e. $u_{x} + u_{y} = 2 (x + y) u$ using method of separation of variables.

Let,

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

\frac{\partial u}{\partial x} = X^{'} \cdot Y & \frac{\partial u}{\partial y} = X \cdot Y^{'}

Now,

\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 2 (x + y) X \cdot Y

\Rightarrow X^{'} \cdot Y + X \cdot Y^{'} = 2 (x + y) X \cdot Y

Dividing by X∙Y both the sides,

\Rightarrow \frac{X^{'} \cdot Y}{X \cdot Y} + \frac{X \cdot Y^{'}}{X \cdot Y} = 2 (x + y)

\Rightarrow \frac{X^{'}}{X} - 2 x = 2 y - \frac{Y^{'}}{Y} = k

∎ \frac{X^{'}}{X} - 2 x = k \Rightarrow \frac{1}{X} dX = (k + 2 x) dx \Rightarrow \log X = kx + x^{2} + c \Rightarrow X = c_{1} e^{kx + x^{2}}

∎ 2 y - \frac{Y^{'}}{Y} = k \Rightarrow \frac{1}{Y} dY = (2 y - K) dy \Rightarrow \log Y = y^{2} - ky + c \Rightarrow Y = c_{1} e^{y^{2} - ky}

Solution:

u (x, y) = X \cdot Y = (c_{1} e^{kx + x^{2}}) (c_{2} e^{y^{2} - ky})

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