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Second order differential equations FP3 AQA question watch

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    I have attached the question in the post. Here is what I have done so far.


    Let  y=e^{mx}

    And I got the auxiliary equation of m^2 -4m +3=0
    Therefore m=1, 3

    So the complimentary function is y=Ae^{x}+Be^{3x}

    So the overall general solution would be y=Ae^{x}+Be^{3x}+Cx+D

    I just don't know how to get the particular integral and particular solution. Help please?
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    (Original post by CharlieBoardman)
    I have attached the question in the post. Here is what I have done so far.


    Let  y=e^{mx}

    And I got the auxiliary equation of m^2 -4m +3=0
    Therefore m=1, 3

    So the complimentary function is y=Ae^{x}+Be^{3x}

    So the overall general solution would be y=Ae^{x}+Be^{3x}+Cx+D

    I just don't know how to get the particular integral and particular solution. Help please?
    Let our particular integral be of the form y=\lambda x+\mu (1) (since our differential equation is equal to 6x-5) where \lambda, \mu are constants. Also your C.F is just y=Ae^{x}+Be^{3x} work separately and then combine the particular integral and complimentary function to gain a general solution

    Differentiate this expression (1) once, and then again to then be plugged back into our initial differential equation to find out the values of \lambda and  \mu
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    (Original post by CharlieBoardman)
    I have attached the question in the post. Here is what I have done so far.


    Let  y=e^{mx}

    And I got the auxiliary equation of m^2 -4m +3=0
    Therefore m=1, 3

    So the complimentary function is y=Ae^{x}+Be^{3x}

    So the overall general solution would be y=Ae^{x}+Be^{3x}+Cx+D

    I just don't know how to get the particular integral and particular solution. Help please?
    To find the particular integral you should let y=Cx+D and substitute into the original differential equation.

    Then to find your particular solution you should simply apply the boundary conditions to your solution and d/dx of your solution.
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    (Original post by CharlieBoardman)
    I have attached the question in the post. Here is what I have done so far.


    Let  y=e^{mx}

    And I got the auxiliary equation of m^2 -4m +3=0
    Therefore m=1, 3

    So the complimentary function is y=Ae^{x}+Be^{3x}

    So the overall general solution would be y=Ae^{x}+Be^{3x}+Cx+D

    I just don't know how to get the particular integral and particular solution. Help please?
    I hope this will help you.
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  1. File Type: pdf Particular Integrals.pdf (16.8 KB, 154 views)
 
 
 
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