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Math Help - boundary value problem

  1. #1
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    Lightbulb boundary value problem

    I had discussed that the boundary value problem y"+y=0 with boundary conditions y'(0)=1 and y(pi/2) = 0 yesterday.

    There was no solution to this because there would be 2 different values for the arbitrary constant c2.

    Now, what change would take place if we had y" +y= x? and the same boundary conditions?

    Will it still have no solution ? The particular solution, in that case, will be of the form y=a+bx ... When I apply the boundary conditions, I see that the a is -pi/2 and b is 1.

    so, does this make the solution be y = x-pi/2? I also see that this merely fails to satisfy the equation given.

    I sense that i'm overlooking something...
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  2. #2
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    Re: boundary value problem

    What is your general solution?
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  3. #3
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    Re: boundary value problem

    Quote Originally Posted by MAX09 View Post
    I had discussed that the boundary value problem y"+y=0 with boundary conditions y'(0)=1 and y(pi/2) = 0 yesterday.

    There was no solution to this because there would be 2 different values for the arbitrary constant c2.

    Now, what change would take place if we had y" +y= x? and the same boundary conditions?

    Will it still have no solution ? The particular solution, in that case, will be of the form y=a+bx ... When I apply the boundary conditions, I see that the a is -pi/2 and b is 1.

    This is wrong.

    so, does this make the solution be y = x-pi/2? I also see that this merely fails to satisfy the equation given.

    I sense that i'm overlooking something...
    .
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    Re: boundary value problem

    some more light on this would be helpful... Isn't the general solution forproblems of this kind like y = y_h + y_p where y_h is the solution to the homogenous equation y"+y=0 and y_p of the form a+bx because the function given equation is 'x'?

    I remember reading that in the method of undetermined coefficients for solving L(y)= f(x), the particular solution,
    y_p = an nth degree polynomial if the degree of f(x) is n.

    @sudharaka... I too have the same intuition.. can you lead me out of this incorrect approach?
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  5. #5
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    Re: boundary value problem

    Quote Originally Posted by MAX09 View Post
    some more light on this would be helpful... Isn't the general solution forproblems of this kind like y = y_h + y_p where y_h is the solution to the homogenous equation y"+y=0 and y_p of the form a+bx because the function given equation is 'x'?

    I remember reading that in the method of undetermined coefficients for solving L(y)= f(x), the particular solution,
    y_p = an nth degree polynomial if the degree of f(x) is n.

    @sudharaka... I too have the same intuition.. can you lead me out of this incorrect approach?
    The particular solution is obviously y=x. It could be also obtained using D operator method.
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    Re: boundary value problem

    @sudharaka-I get the answer by the D operator method. but why is the method of undetermined coefficients not working? in this case, since the polynomial function on the right is of degree one, my particular solution has got to be one where i have a constant and an 'x' term. the constant should automatically vanish leaving way to the 'x' alone. why isn't this happening here????

    kindly bear with me on this one...
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  7. #7
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    Re: boundary value problem

    Quote Originally Posted by MAX09 View Post
    @sudharaka-I get the answer by the D operator method. but why is the method of undetermined coefficients not working? in this case, since the polynomial function on the right is of degree one, my particular solution has got to be one where i have a constant and an 'x' term. the constant should automatically vanish leaving way to the 'x' alone. why isn't this happening here????

    kindly bear with me on this one...
    Hi MAX09,

    If y=a+bx is your guess for the particular integral it should satisfy the differential equation and hence you should be able to find the two coefficients a and b.

    y=a+bx

    \Rightarrow \frac{d^{2}y}{dx^2}=0

    \mbox{Therefore, }\frac{d^{2}y}{dx^2}+y=a+bx-------(1)

    The given differential equation is, \frac{d^{2}y}{dx^2}+y=x--------(2)

    By comparing the right hand sides of equation (1) and (2); a=0 and b=1. Hence the particular integral is, y=a+bx=0+(1)x=x.
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  8. #8
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    Re: boundary value problem

    Quote Originally Posted by MAX09 View Post
    some more light on this would be helpful... Isn't the general solution forproblems of this kind like y = y_h + y_p where y_h is the solution to the homogenous equation y"+y=0 and y_p of the form a+bx because the function given equation is 'x'?

    I remember reading that in the method of undetermined coefficients for solving L(y)= f(x), the particular solution,
    y_p = an nth degree polynomial if the degree of f(x) is n.

    @sudharaka... I too have the same intuition.. can you lead me out of this incorrect approach?
    I meant to ask: what is your general solution for this particular DE?
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