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    Hi,

    I don't understand why we write the general solution of second order differential equations as the sum of the two solutions.

    i.e.
    (d^2y/dx^2) + 5(dy/dx) + 6y = 0
    Let y = e^(mx) etc.

    Eventually I come to the point where I say:
    m = -2 and m=-3
    So therefore, y = e^(-2x) and y = e^(-3x)

    My question is, why is the solution written as y = Ae^(-2x) + B^e^-(3x) ?

    Thanks in advance. The example above was taken directly from the Edexcel textbook.
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    If  f(x) is a solution to the ODE then it's easy to show that  Af(x) is also a solution, the same goes for showing that  Bg(x) is a solution if  g(x) is.
    Then from there it is also easy to show that  Af(x) + Bg(x) is also a solution.
    I suggest you try it for yourself.
    Is you have  y''-y'+y=0 .
    If  f(x) is a solution then clearly  f''(x)-f'(x)+f(x)=0 .
    Then if we let  y=Af(x) , then we have  Af''(x)-Af'(x)+Af(x)\equiv A(f''(x)-f'(x)+f(x)) and this clearly is 0. So it too is a solution. You can similarly prove all the other results I said.
    You can prove this more generally using the same method.
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    (Original post by B_9710)
    If  f(x) is a solution to the ODE then it's easy to show that  Af(x) is also a solution, the same goes for showing that  Bg(x) is a solution if  g(x) is.
    Then from there it is also easy to show that  Af(x) + Bg(x) is also a solution.
    I suggest you try it for yourself.
    Is you have  y''-y'+y=0 .
    If  f(x) is a solution then clearly  f''(x)-f'(x)+f(x)=0 .
    Then if we let  y=Af(x) , then we have  Af''(x)-Af'(x)+Af(x)\equiv A(f''(x)-f'(x)+f(x)) and this clearly is 0. So it too is a solution. You can similarly prove all the other results I said.
    You can prove this more generally using the same method.
    I tried it out and of course you are correct. Thanks for all your help and time!
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    (Original post by JLegion)
    I tried it out and of course you are correct. Thanks for all your help and time!
    And because it's second order, there must be two constants in the solution. Finding the solution is equivalent to integrating twice, which would give you two constants.
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    (Original post by tiny hobbit)
    And because it's second order, there must be two constants in the solution. Finding the solution is equivalent to integrating twice, which would give you two constants.
    Yes, thank you
 
 
 
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