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Math Help - Need help solving some 2nd order DEs

  1. #1
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    Need help solving some 2nd order DEs

    Hi to everyone. I need help. I have few equations that i can not solve. If there is anyone that can help...please solve them step by step. I need them as fast as posible.....


    1. y"+y'=e^x+2*cosx
    2. x^2*y"-2*x*y'+2*y=2*x^3-x
    3. y"+y'-2*y=(x^2-1)e^(2*x)
    4. y"'-4y'=x*e^(2*x)+sinx
    5. What kind of differential equation will we have if we put x=1/t in x^4*y"+2*x^3*y'-4*y=1/x (the new differential equation should be solved using Laplas transforamtion)

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  2. #2
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    OK, Let me explain the second one in detail, the other ones follow the same method (variation of parameters). First the substitution needed is [imath]x=e^t[/imath]. This is only for the second one, not for the last equation. I did not solve the others, presumably they can be found also by variation of parameters. The substitution is thus:
    x=e^t
    from which
    dx=e^tdt
    and thus:
    \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=e^{-t}\frac{dy}{dt}
    \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\  right)= \frac{dt}{dx}\frac{d}{dt}\left(e^{-t}\frac{dy}{dt}\right)=e^{-2t}\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)
    Substituting this in the DE gives:
    y''-3y'+2y=2e^{3t}-e^t
    The solution to the homogeneous equation is found as:
    y=Ae^t+Be^{2t}
    Variation of parameters is now entering the picture. I find the easiest way to apply this method is by taking the derivative of the homogeneous solution that many times as one less than the order. This means here that we have a second order equation, so we need one less or only a first derivative. If the equation was a fourth order one we would need three times the derivative. It all will become clear. So the derivative of the solution is (treating A and B as a constant):
    y'=Ae^t+2Be^{2t}
    This together with the original solution gives a system of two equations in two unknowns (A and B), but must be written correctly. The latter meaning that you need to set all the derivatives equal to zero except the last one, this must be set equal to the RHS of the DE. Furthermore you need to rewrite the A and B as A' and B', and solve for these. After you have solved them you have the derivative of A and B and two integrations are all that is left to find them. After that you need to substitute them back into the solution and you have the solution to the complete equation. Let's do this. The system is:
    A'e^t+B'e^{2t}=0
    A'e^t+2B'e^{2t}=2e^{3t}-e^t
    This is an algebraic system and can be solved for A' and B' in a standard way, it gives:
    A'=1-2e^{2t}
    B'=2e^t-e^{-t}
    All we need to do is to integrate, giving:
    A=-e^{2t}+t+K_1
    B=2e^t+e^{-t}+K_2
    Putting this into the original solution gives:
    y=-e^{3t}+te^t+K_1e^t+2e^{3t}+e^t+K_2e^{2t}
    or:
    y=(K_1+t)e^t+K_2e^{2t}+e^{3t}
    Finally we need to inverse substitute the first substitution, giving:
    y=(K_1+ln(x))x+K_2x^2+x^3
    And tatata, the solution to equation two. You can verify this by substituting the solution into the DE.

    Hope this helps,

    coomast
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  3. #3
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    Quote Originally Posted by Birac View Post
    Hi to everyone. I need help. I have few equations that i can not solve. If there is anyone that can help...please solve them step by step. I need them as fast as posible.....


    1. y"+y'=e^x+2*cosx
    2. x^2*y"-2*x*y'+2*y=2*x^3-x
    3. y"+y'-2*y=(x^2-1)e^(2*x)
    4. y"'-4y'=x*e^(2*x)+sinx
    5. What kind of differential equation will we have if we put x=1/t in x^4*y"+2*x^3*y'-4*y=1/x (the new differential equation should be solved using Laplas transforamtion)

    my e-mail:
    removed by CB


    Thanks
    These appear to come from an assignment that counts towards your final grade. You can disuss this with me via pm. Thread closed.
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