Mumbai University > Information Technology, Computer Engineering > Sem 3 > Applied Mathematics 3

Marks: 5M

Year: Dec 2013

Differentiating the given relation w . r . t x and y

$3 \frac{∂u}{∂x} + 2 {∂v}{∂x} = -2x + 16y$ ---------- (i)

And

$3 {∂u}{∂y} + 2\frac{∂v}{∂y} = 2y + 16x$ ----------- (ii)

But ux = vy and uy = -vx

Hence from (ii) we get

$-3 \frac{∂v}{∂x} + 2 \frac{∂u}{∂x} = 2y + 16x$ -------------- (iii)

Now multiply (i) by (3) and (iii) by (2) and add

$:. 13\frac{∂u}{∂x} = 26x + 52y i.e, \frac{∂u}{∂x} = 2x + 4y = φ1(x,y)$

Again , multiply (i) by (-2) and (iii) by (3) and add

$:. -13 \frac{∂v}{∂x} = 56x – 26y i.e \frac{∂v}{∂x} = -4x + 2y = φ2(x,y)$

But $f^1(z) = \frac{∂u}{∂x} + i\frac{∂v}{∂x} = φ1(x,y) + i φ2(x,y)$

Now making use of Milne Thompson i.e replacing x with z and y with 0 we get = φ1(z ,0) + i φ2(z,0)

$:. f^1(z) = 2z – i. 4z$

Integrating both sides we get

$f(z)= 2 \frac{z^2}{2} – i \frac{4z^2}{2} +c\\ \hspace{0.8cm} = z^2 – 2iz^2 + c\\ \hspace{0.8cm} = z^2(1- 2i ) + c$