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Math Help - differential equation using the method of undetermined coefficient coefficients

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    differential equation using the method of undetermined coefficient coefficients

    Solve the following differential equation using the method of undetermined coefficient coefficients

    d^2y/dx^2 + 9y = x^2 cos3x

    This was the most difficult assignments i had encounted as this was not taught in our module yet we are required to solve it . what i had learn before are the right hand side x^2 + cos3x but not multiply .

    r^2 + 9 = 0
    r(r+ 9) = 0
    r=+/- 3j
    Yc =C1cos3x + C2sin3x

    Yp= (Ax^2 + Bx + C) (Dcos3x + Esin3x) ?
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    Quote Originally Posted by higoodhi View Post
    Solve the following differential equation using the method of undetermined coefficient coefficients

    d^2y/dx^2 + 9y = x^2 cos3x

    This was the most difficult assignments i had encounted as this was not taught in our module yet we are required to solve it . what i had learn before are the right hand side x^2 + cos3x but not multiply .

    r^2 + 9 = 0
    r(r+ 9) = 0
    r=+/- 3j
    Yc =C1cos3x + C2sin3x

    Yp= (Ax^2 + Bx + C) (Dcos3x + Esin3x) ?
    Almost. Since cos 3x and sin 3x are already solutions to the homogeneous equation, multiply that polynomial by x:
    Y_p= (Ax^3+ Bx^2+ Cx)(D cos(3x)+ E sin(3x)).
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    why is it Yp= (Ax^3 + Bx^2 + Cx) (Dcos3x + Esin3x) n not Yp= (Ax^2 + Bx + C) (Dcos3x + Esin3x) ?


    to solve the equation we need to find yp,yp1,yp2
    yp1 differential yp by dx , and yp2 is differential by yp1.

    how to solve it,since it will become very complex as Ax^3.Dcos3x =ADx^3cos 3x(2 unkown is inside AD), Ax^3.Esin3x =AEx^sin 3x(2 unkown is inside AE)......... (Rem this is only yp ) still need to solve yp1 and yp2 there will be a lot of unknown .....

    is there any simple method ?

    aft substutite Yp2 and yp into d^2y/dx^2 + 9y = x^2 cos3x

    where we need to find the value of A,B,C,D,E
    Last edited by higoodhi; March 14th 2009 at 05:33 AM. Reason: more precise
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by higoodhi View Post
    why is it Yp= (Ax^3 + Bx^2 + Cx) (Dcos3x + Esin3x) n not Yp= (Ax^2 + Bx + C) (Dcos3x + Esin3x) ?
    you can't have your particular solution of the same form as your homogeneous solution. you will just get zero when you plug everything in, which gets you nowhere. to avoid this, multiplying through by x is the standard trick.

    to solve the equation we need to find yp,yp1,yp2
    yp1 differential yp by dx , and yp2 is differential by yp1.
    no, you need to find yh (the homogeneous solution) and yp. that's it

    how to solve it,since it will become very complex as Ax^3.Dcos3x =ADx^3cos 3x(2 unkown is inside AD), Ax^3.Esin3x =AEx^sin 3x(2 unkown is inside AE)......... (Rem this is only yp ) still need to solve yp1 and yp2 there will be a lot of unknown .....

    is there any simple method ?

    aft substutite Yp2 and yp into d^2y/dx^2 + 9y = x^2 cos3x

    where we need to find the value of A,B,C,D,E
    what you typed is confusing. anyway, the method of undetermined coefficients says to find y_p, y_p' and y_p'' and plug them into your original differential equation. equate the coefficients to solve for A, B, C, D and E
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    Quote Originally Posted by Jhevon View Post
    anyway, the method of undetermined coefficients says to find y_p, y_p' and y_p'' and plug them into your original differential equation. equate the coefficients to solve for A, B, C, D and E
    how to solve the product unkown since the product unkown will become very complex as Ax^3.Dcos3x =ADx^3cos 3x(2 unkown is inside AD), Ax^3.Esin3x =AEx^sin 3x(2 unkown is inside AE)......
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by higoodhi View Post
    how to solve the product unkown since the product unkown will become very complex as Ax^3.Dcos3x =ADx^3cos 3x(2 unkown is inside AD), Ax^3.Esin3x =AEx^sin 3x(2 unkown is inside AE)......
    leave off D and E. just forget about those. solve for A, B and C
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    Quote Originally Posted by Jhevon View Post
    leave off D and E. just forget about those. solve for A, B and C
    so do u mean
    Yp= (Ax^3 + Bx^2 + Cx) (cos3x + sin3x) ?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by higoodhi View Post
    so do u mean
    Yp= (Ax^3 + Bx^2 + Cx) (cos3x + sin3x) ?
    yes
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    thank
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