Complex Variables and Partial Differential Equations (3130005)

Question-4c-OR

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

Q4) (c)

Find the general solution of partial differential equation $u_{xx} = 9 u_{y}$ using method of separation of variables.

→ Let,

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

\Rightarrow \frac{\partial^{2} u}{\partial x^{2}} = u_{xx} = X^{''} \cdot Y & \frac{\partial u}{\partial y} = u_{y} = X \cdot Y^{'}

→ Now,

u_{xx} = 9 u_{y}

\Rightarrow X^{''} \cdot Y = 9 X \cdot Y^{'}

→ Dividing by

X \cdot Y

both the sides,

\Rightarrow \frac{X^{''} \cdot Y}{X \cdot Y} = 9 \frac{X \cdot Y^{'}}{X \cdot Y}

\Rightarrow \frac{X^{''}}{X} = 9 \frac{Y^{'}}{Y} = K (Consider, K > 0)

\Rightarrow \frac{X^{''}}{X} = K & 9 \frac{Y^{'}}{Y} = K

∎ \frac{X^{''}}{X} = K

\Rightarrow X^{''} - KX = 0

→ Auxiliary equation is

m^{2} - K = 0 \Rightarrow m = \pm \sqrt{K}

\Rightarrow X = c_{1} e^{\sqrt{K}} + c_{2} e^{- \sqrt{K}}

∎ \frac{Y^{'}}{Y} = K

\Rightarrow \frac{1}{Y} \frac{dY}{dy} = \frac{K}{9}

\Rightarrow \frac{1}{Y} dY = \frac{K}{9} dy

\Rightarrow \int \frac{1}{Y} dY = \int \frac{K}{9} dy \Rightarrow \log Y = \frac{Ky}{9} + C \Rightarrow Y = e^{\frac{Ky}{9} + C} \Rightarrow Y = c_{3} e^{\frac{Ky}{9}}

Solution:

u (x, y) = X \cdot Y = (c_{1} e^{\sqrt{K}} + c_{2} e^{- \sqrt{K}}) (c_{3} e^{\frac{Ky}{9}})

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