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    Okay, I'm trying to find the general solution of a second ODE, but I'm finding a problem with the particular integral.

    \frac{d^2y}{dx^2}-2\frac{dy}{dx}+26y=e^x

    Worked out the complementary function as y=e^x(Acos5x+Bsin5x) which is correct. Now here's my method for the particular integral:

    Let y=\lambda e^x. However, that particular integral is on the complementary function, so I know I cannot use it. Therefore, I have decided to let y=\lambda xe^x. From here I get the following:

    \frac{dy}{dx}=\lambda xe^x+\lambda e^x

    \frac{d^2y}{dx^2}=\lambda xe^x+2\lambda e^x

    \therefore \lambda xe^x+2\lambda e^x-2\lambda xe^x-2\lambda e^x+26\lambda xe^x=e^x

    25\lambda x=1

    However, this obviously presents a problem. The answer should be \lambda =\frac{1}{25}, but I've hit a brick wall here.
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    (Original post by ViralRiver)
    Okay, I'm trying to find the general solution of a second ODE, but I'm finding a problem with the particular integral.

    \frac{d^2y}{dx^2}-2\frac{dy}{dx}+26y=e^x

    Worked out the complementary function as y=e^x(Acos5x+Bsin5x) which is correct. Now here's my method for the particular integral:

    Let y=\lambda e^x. However, that particular integral is on the complementary function, so I know I cannot use it. Therefore, I have decided to let y=\lambda xe^x. From here I get the following:

    \frac{dy}{dx}=\lambda xe^x+\lambda e^x

    \frac{d^2y}{dx^2}=\lambda xe^x+2\lambda e^x

    \therefore \lambda xe^x+2\lambda e^x-2\lambda xe^x-2\lambda e^x+26\lambda xe^x=e^x

    25\lambda x=1

    However, this obviously presents a problem. The answer should be \lambda =\frac{1}{25}, but I've hit a brick wall here.
    I may be wrong, but I think you do not need to multiply your PI by a factor of x. Although e^x does appear, it appears multiplied to the cos and sin functions. You would only need to multiply through by your factor of x, if you had some constant multiplied by it i.e. Ae^x or Be^x
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    dknt: correct.
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    Ok, I'm confused >< . Why does it make a difference if it's a product of e^x and sin/cosine? Also, shouldn't it work out in the end regardless of whether you have to multiply by x or not?
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    (Original post by ViralRiver)
    Ok, I'm confused >< . Why does it make a difference if it's a product of e^x and sin/cosine? Also, shouldn't it work out in the end regardless of whether you have to multiply by x or not?
    Well, no it shouldn't work out, because an x term has just popped up in your solution, when it shouldn't.

    Your PI will be in the form of \lambda e^x. This term does not appear in your CF and so can be used. From your CF and multiplying out,

    Ae^x\mathrm{cos}5x + Be^x\mathrm{cos}5x

    As you can see an A e^x or B e^x, solely, does not occur. If they did occur, then you would need to multiply through by a factor of x in your PI.
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    Okay, then I'd like an explanation on how to know when I can't use a standard particular integral, if possible? The book I'm reading says you cannot use it if it is a factor of the CF, which in this case it is =\ .
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    (Original post by ViralRiver)
    Okay, then I'd like an explanation on how to know when I can't use a standard particular integral, if possible? The book I'm reading says you cannot use it if it is a factor of the CF, which in this case it is =\ .
    Okay, so let's say you have the DE \dfrac{d^2y}{dx^2}-3 \dfrac{dy}{dx} =6

    Working through it you end up with a CF of,

    y=A +Be^{3x}

    Now usually, you PI will be in the form of,

    y= \lambda, where lambda is a constant. However, if we look at our CF, you can see there is already a constant A. Since your general solution is y=CF+PI, you cannot use y= \lambda and so you must have a factor of x in there i.e. y= \lambda x

    EDIT

    Another,

    \dfrac{d^2y}{dx^2}-4 \dfrac{dy}{dx} +4y =e^{2x}

    Your CF ends up to be y=(A+Bx)e^{2x}

    Your PI would usually be,

    y=\lambda e^{2x}, but y=A e^{2x} already appears.

    You could try y=\lambda x e^{2x}, but,

    y=Bx e^{2x} also appears, so you must use,

    y=\lambda x^2 e^{2x}
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    Ok, that makes a lot of sense thanks . Just one query, is the CF in your first example wrong, or am I missing something? Surely m^2-3m-6=0 has different roots to what you have stated?

    Also, why can't such a particular integral be used, what I mean is, what effect does it have, being included in the CF?
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    (Original post by ViralRiver)
    Ok, that makes a lot of sense thanks . Just one query, is the CF in your first example wrong, or am I missing something? Surely m^2-3m-6=0 has different roots to what you have stated?

    Also, why can't such a particular integral be used, what I mean is, what effect does it have, being included in the CF?
    In answer to the bold, your Auxiliary Quadratic Equation is m^2 -3m=0, since there is no y term. The 6 is on the RHS

    In terms of "why" it can't be used, I don't know the reason, other than it just doesn't work out
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    (Original post by dknt)
    In answer to the bold, your Auxiliary Quadratic Equation is m^2 -3m=0, since there is no y term. The 6 is on the RHS

    In terms of &quot;why&quot; it can't be used, I don't know the reason, other than it just doesn't work out
    Okay, thanks a lot for all your help .


    (Original post by DFranklin)
    dknt: correct.
    Would you happen to know why particular integrals can't be used when contained within the CF itself?
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    (Original post by ViralRiver)
    Would you happen to know why particular integrals can't be used when contained within the CF itself?
    Think about what the definition of the complementary function is, and compare it to the definition of . Each term in the complementary function causes the LHS to evaluate to zero. If your particular integral is a term in the complementary function, then the LHS will evaluate to zero rather than the function on the RHS of the equation, as a particular integral should.
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    (Original post by Farhan.Hanif93)
    Think about what the definition of the complementary function is, and compare it to the definition of . Each term in the complementary function causes the LHS to evaluate to zero. If your particular integral is a term in the complementary function, then the LHS will evaluate to zero rather than the function on the RHS of the equation, as a particular integral should.
    Okay, thanks .

    One more question. When finding the complementary function, from, for example \frac{d^2y}{dx^2}-y=2e^x is it alright to work it out by sight, i.e m^2-1=0 etc? I ask this as the textbook says to let y=e^{mx}, \frac{d^2y}{dx^2}=m^2e^{mx} then cancelling out the e^{mx} as e^{mx}&gt;0. On that note, if e^{mx}&lt;0 (I know it can't be), would dividing by it not work or something?
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    (Original post by ViralRiver)
    Okay, thanks .

    One more question. When finding the complementary function, from, for example \frac{d^2y}{dx^2}-y=2e^x is it alright to work it out by sight, i.e m^2-1=0 etc? I ask this as the textbook says to let y=e^{mx}, \frac{d^2y}{dx^2}=m^2e^{mx} then cancelling out the e^{mx} as e^{mx}&gt;0. On that note, if e^{mx}&lt;0 (I know it can't be), would dividing by it not work or something?
    Yes, using the auxiliary equation is fine.

    As for your other question, it depends on the type of equality you're dealing with.
    If it's a case of solving a quadratic equation, for example, then the real problem arises from the risk of dividing by zero, and thus losing solutions. It is not a problem to divide or multiply by a quantity that is negative.

    In an inequality, there's the added worry of what happens when you multiply of divide through by a negative. If you were to multiply or divide by a quantity less than zero, you would need to flip the inequality sign to adjust for it otherwise the statement will be wrong.
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    (Original post by Farhan.Hanif93)
    Yes, using the auxiliary equation is fine.

    As for your other question, it depends on the type of equality you're dealing with.
    If it's a case of solving a quadratic equation, for example, then the real problem arises from the risk of dividing by zero, and thus losing solutions. It is not a problem to divide or multiply by a quantity that is negative.

    In an inequality, there's the added worry of what happens when you multiply of divide through by a negative. If you were to multiply or divide by a quantity less than zero, you would need to flip the inequality sign to adjust for it otherwise the statement will be wrong.
    Ok thanks .

    Something's popped up now which doesn't really make sense either =\ .

    \frac{d^2x}{dt^2}-4\frac{dx}{dt}+4x=3te^{2t}

    Now, for the particular integral, I would use \lambda t^2e^{2t} as that does not appear in the CF. However, the solution requires you to use \lambda t^3e^{2t} oO. Seems like I really don't understand it after all :< .
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    (Original post by ViralRiver)
    Ok thanks .

    Something's popped up now which doesn't really make sense either =\ .

    \frac{d^2x}{dt^2}-4\frac{dx}{dt}+4x=3te^{2t}

    Now, for the particular integral, I would use \lambda t^2e^{2t} as that does not appear in the CF. However, the solution requires you to use \lambda t^3e^{2t} oO. Seems like I really don't understand it after all :< .
    That's odd. Have you tried it with the t^2 term? If so, what happened?
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    (Original post by Farhan.Hanif93)
    That's odd. Have you tried it with the t^2 term? If so, what happened?
    I haven't directly tried it, but the book specifically pointed me in that direction (check attachment).
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    (Original post by ViralRiver)
    I haven't directly tried it, but the book specifically pointed me in that direction (check attachment).
    I'm not too sure. I suggest that you should work it through and see why it doesn't work.

    Hopefully someone else can take over from here, I'm not the most experienced with ODEs.
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    (Original post by ViralRiver)
    I haven't directly tried it, but the book specifically pointed me in that direction (check attachment).
    I can only guess that perhaps it's a "special case" and they've given you the PI to use, because it goes against the usual method, for some reason. Or it's a typo.... But I doubt it's the latter.
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    (Original post by Farhan.Hanif93)
    I'm not too sure. I suggest that you should work it through and see why it doesn't work.

    Hopefully someone else can take over from here, I'm not the most experienced with ODEs.

    (Original post by dknt)
    I can only guess that perhaps it's a &quot;special case&quot; and they've given you the PI to use, because it goes against the usual method, for some reason. Or it's a typo.... But I doubt it's the latter.
    Oh wow... using that reduces to 2\lambda =3t. Very weird oO. If anyone has an explanation I'd love to know .
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    (Original post by dknt)
    I can only guess that perhaps it's a "special case" and they've given you the PI to use, because it goes against the usual method, for some reason. Or it's a typo.... But I doubt it's the latter.
    Couldn't actually remember how to solve ODEs 'normally', but worked it through using Laplace Transforms and there's definitely a t^3 term in the answer, so definitely not a typo. (Yes, I know you said you doubt it; just confirming).

    I can't think of any reason why though...would be most intrigued as to the answer if anyone here knows it!
 
 
 
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