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Math Help - solve the system through systematic elimination...

  1. #1
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    solve the system through systematic elimination...

    ok im not sure if im doing this right, my numbers are getting a little funky i was wondering if someone could check my work and let me know if im doing it right. this is the problem



    i solved it for y by eliminating the x

    i got the following...



    im not done with the problem, i jus wanted to know if i was right up to the point of where I am. thanks in advance.
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  2. #2
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    D is operator? If the problem is like that you shouldn't separate from variable!
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  3. #3
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    Yes, "D" is the derivative operator. But it is common notation in linear differential equations with constant coefficients to write, say, \frac{d^2y}{dt^2}+ 5\frac{dy}{dt}+ 7y= 0 as (D^2+ 5D+ 7)y= 0. As long as the coefficients are constant, the " D^2+ 5D+ 7" can be treated like a polynomial (if the coefficients are not constant, you run into problems with commutativity).

    Slapmaxwell1, that is not what I get. Differentiating the second equation, (D^2- 9D)y- 2Dx= 0. From the first equation, Dx= 5x- 7y so (D^2- 9D)y- 2(5x- 7y)= (D^2- 9D+ 14)y- 10x= 0. From the second equation, [tex]2x= (D- 9)y[tex] so [tex]-10x= (-5D+ 45)y. Then (D^2- 9D+ 14)y- 10x= (D^2- 9D+ 14)y+ (-5D+ 45)y= (D^2- 14D+ 59)y= 0. You appear to have dropped the "14y".

    D^2- 14D+ 59 is not going to factor easily but the equation D^2- 14D+ 59= 0 can be solved by completing the square or the quadratic formula.
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  4. #4
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    I didn't see. You are right in your world. But in my, completely wrong. You didn't finish a problem (in your world). I will get solutions always before you in my world.
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  5. #5
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    ok i will rework the problem thanks alot for your help!
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    Yes, "D" is the derivative operator. But it is common notation in linear differential equations with constant coefficients to write, say, \frac{d^2y}{dt^2}+ 5\frac{dy}{dt}+ 7y= 0 as (D^2+ 5D+ 7)y= 0. As long as the coefficients are constant, the " D^2+ 5D+ 7" can be treated like a polynomial (if the coefficients are not constant, you run into problems with commutativity).

    Slapmaxwell1, that is not what I get. Differentiating the second equation, (D^2- 9D)y- 2Dx= 0. From the first equation, Dx= 5x- 7y so (D^2- 9D)y- 2(5x- 7y)= (D^2- 9D+ 14)y- 10x= 0. From the second equation, [tex]2x= (D- 9)y[tex] so [tex]-10x= (-5D+ 45)y. Then (D^2- 9D+ 14)y- 10x= (D^2- 9D+ 14)y+ (-5D+ 45)y= (D^2- 14D+ 59)y= 0. You appear to have dropped the "14y".

    D^2- 14D+ 59 is not going to factor easily but the equation D^2- 14D+ 59= 0 can be solved by completing the square or the quadratic formula.
    thanks so much for your help, i really appreciate it, i will rework the problem with the suggested correction.
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  7. #7
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    Regards
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