Poly Engineering MAths year 2 (Differential Equations)

• Wireless-

  7 Jul 10, 00:13

  2b) Fine the particular solutions of the following differential equations

   

  (4x+xy^2)dx +(y+yx^2)dy=0 , given that y(1)=2

  Ans: (4+y^2)(1+x^2)=16

   

   

  how to do. I cant seperate the y and the x variables in the equations. Help

• -Wanderer-

  7 Jul 10, 02:10

  You learnt Exact Differential Equations before?

  They are in the form of: M(x,y) dx + N(x,y) dy = 0.

  Let M = 4x + xy^2,        N = y + yx^2

  Partial diff M w.r.t. y to get 2xy

  Partial diff N w.r.t. x to get 2xy as well. 

  Therefore, this proves that the equation is exact.

  Let a function f(x,y) = k such that:

  (del f / dx) = M = 4x + xy^2     - (1)

  (del f / dy) = N = y + yx^2        - (2)

  Integrate (del f / dx) w.r.t. x to get:

  f(x,y) = 2x^2 + 0.5x^2(y^2) + g(y)    - (3)

  Partial differentiate f(x,y) w.r.t. y to get

  0 + yx^2 + g'(y)

  Compare this with (2), you will see that

  g'(y) = y

  g(y) = 0.5y^2 + c

  Sub g(y) into (3)

  f(x,y) = 2x^2 + 0.5x^2(y^2) + 0.5y^2 + c

  k - c = 2x^2 + 0.5x^2(y^2) + 0.5y^2

  C = 2x^2 + 0.5x^2(y^2) + 0.5y^2    where C = k - c

  When x = 1, y = 2,

  C = 6

  6 = 2x^2 + 0.5x^2(y^2) + 0.5y^2

  12 = 4x^2 + x^2(y^2) + y^2

  16 = 4 + 4x^2 + x^2(y^2) + y^2

  16 = (4+y^2)(1+x^2)

  ________________________________________________________________

  And I just realised that my method is too unorthodox. Lol! Follow it if you want to. Cheers! =D

• wee_ws

  7 Jul 10, 08:48

  Hi,

    It takes some careful manipulation to enable us to use variable separable:

    dy/dx = -{x(4 + y^2)} / {y(1 + x^2)}

    dy/dx = -{x/(1 + x^2)} . {(4 + y^2)/y},
    where we group expressions involving x, y separately.

    Now we can apply variable separable:

    integral y/(4 + y^2) dy = - integral x/(1 + x^2) dx, from which you should be able to proceed on your own.

    Thanks.

  Cheers,
  Wen Shih

• eagle

  7 Jul 10, 08:51

  Looks good for A levels as well.

  Thanks for sharing :)

• Wireless-

  7 Jul 10, 11:53

  Thanks wee_ws. Got it!

   

  • Reply/Quote

  Wanderer we have not learned Exact Differential Equations.

• -Wanderer-

  8 Jul 10, 20:47

  Originally posted by wee_ws:

  Hi,

    It takes some careful manipulation to enable us to use variable separable:

    dy/dx = -{x(4 + y^2)} / {y(1 + x^2)}

    dy/dx = -{x/(1 + x^2)} . {(4 + y^2)/y},
    where we group expressions involving x, y separately.

    Now we can apply variable separable:

    integral y/(4 + y^2) dy = - integral x/(1 + x^2) dx, from which you should be able to proceed on your own.

    Thanks.

  Cheers,
  Wen Shih

  
  I couldn't spot that pattern. Very good one! =D

• wee_ws

  8 Jul 10, 22:48

  Hi Wanderer,

    Thanks :)

    Initially I thought it followed the homogeneous form. After a failed attempt with the substitution y = vx and Wireless' hint of variable separable, I decided to scrutinise the complex expression and saw through its trickery :P

    The setter is ingenious to think of something like that!

  Cheers,
  Wen Shih