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Math Help - First order non-linear ODE

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    First order non-linear ODE

    Hi everyone,

    x2y'-20+x2y2=0; y(1)=1

    The question asks me to use substitution

    y=1/u*du/dx to solve

    So far I have

    y'=-1/u2*du/dx+1/u*d2u/dx2

    problem is when I sub in for the x2y2 term I'm going to get a derivative squared --> y2=(1/u*du/dx)2. Don't know how to resolve this. Please help.
    Last edited by lillybeans; October 16th 2012 at 04:33 PM.
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    Re: First order non-linear ODE

    Quote Originally Posted by lillybeans View Post
    Hi everyone,

    x2y'-20+x2y2=0; y(1)=1

    The question asks me to use substitution

    y=1/u*du/dx to solve

    So far I have

    y'=-1/u2*du/dx+1/u*d2u/dx2

    problem is when I sub in for the x2y2 term I'm going to get a derivative squared --> y2=(1/u*du/dx)2. Don't know how to resolve this. Please help.
    That term should reduce out....

    y=\frac{u'}{u} \iff uy=u'

    y'u+u'y=u'' \iff y'u+u'\frac{u'}{u}=u'' \iff y'=-\frac{(u')^2}{u^2}+u''

    If you sub this into the ode you get

    x^2\left( -\frac{(u')^2}{u^2}+u''\right)+x^2\left(\frac{u'}{u  } \right)^2=20

    This gives

    u''(x)=\frac{20}{x^2}

    Can you finish from here?
    Thanks from topsquark
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    Re: First order non-linear ODE

    Quote Originally Posted by lillybeans View Post
    Hi everyone,

    x2y'-20+x2y2=0; y(1)=1

    The question asks me to use substitution

    y=1/u*du/dx to solve

    So far I have

    y'=-1/u2*du/dx+1/u*d2u/dx2

    problem is when I sub in for the x2y2 term I'm going to get a derivative squared --> y2=(1/u*du/dx)2. Don't know how to resolve this. Please help.
    After simplification, the terme (1/u*du/dx) disapears
    Last edited by JJacquelin; October 16th 2012 at 09:51 PM.
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    Re: First order non-linear ODE

    Thank you both I solved this problem BUT I have another one which I am really stuck on this time, supposedly a much easier problem.

    There are two questions that I am trying to solve on web assignment. The goal is to find a general form of a particular solution to each ODE. The question asks me to represent all constants in the solution using "P,Q,R,S,T..etc.", in that order.

    1. y'''-9y''+14y'=x2
    2. y''-9y'+14y=x2e4x

    For the first one I wrote:
    yp=Px3+Qx2+Rx

    Second one:
    yp=(Px2+Qx+R)e4x

    Neither of the answers are correct, according to the computer. Where did I go wrong?
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    Re: First order non-linear ODE

    Quote Originally Posted by lillybeans View Post
    Thank you both I solved this problem BUT I have another one which I am really stuck on this time, supposedly a much easier problem.

    There are two questions that I am trying to solve on web assignment. The goal is to find a general form of a particular solution to each ODE. The question asks me to represent all constants in the solution using "P,Q,R,S,T..etc.", in that order.

    1. y'''-9y''+14y'=x2
    2. y''-9y'+14y=x2e4x

    For the first one I wrote:
    yp=Px3+Qx2+Rx

    Second one:
    yp=(Px2+Qx+R)e4x

    Neither of the answers are correct, according to the computer. Where did I go wrong?
    I am not sure what method you are using to find the particular solutions, but here is one...

    We can write the ODE as

    (D^3-9D^2+14D)y=x

    Factoring the differential operator we get

    D(D-7)(D-2)y=x

    So to make the equation equal to zero we act on both sides with D^2

    D^3(D-7)(D-2)y=0

    So D^3 annihilates

    y=c_1+c_2x+c_3x^2

    but we must get rid of the part from the complimentary solution (The constant term) so we get

    y=c_2x+c_3x^2

    Let me know if this works for you
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    Re: First order non-linear ODE

    Thank you very much for you help, but unfortunately that didn't work. I even checked my answer using Wolfram and it seems to be correct, maybe something is wrong with web assignment.

    I used the method of undetermined coefficients to find the particular solution.
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    Re: First order non-linear ODE

    Quote Originally Posted by lillybeans View Post
    Thank you both I solved this problem BUT I have another one which I am really stuck on this time, supposedly a much easier problem.

    There are two questions that I am trying to solve on web assignment. The goal is to find a general form of a particular solution to each ODE. The question asks me to represent all constants in the solution using "P,Q,R,S,T..etc.", in that order.

    1. y'''-9y''+14y'=x2
    2. y''-9y'+14y=x2e4x

    For the first one I wrote:
    yp=Px3+Qx2+Rx

    Second one:
    yp=(Px2+Qx+R)e4x

    Neither of the answers are correct, according to the computer. Where did I go wrong?
    For Question 1, rewrite your equation as \displaystyle \begin{align*} u'' - 9u' + 14u = x^2 \end{align*} by letting \displaystyle \begin{align*} u = y' \end{align*}. Then to solve for \displaystyle \begin{align*} u \end{align*}, the homogeneous solution has complementary function

    \displaystyle \begin{align*} m^2 - 9m + 14 &= 0 \\ (m - 2)(m - 7) &= 0 \\ m = 2 \textrm{ or } m = 7 \end{align*}

    So the homogeneous solution is \displaystyle \begin{align*} C_1 \, e^{2x} + C_2 \, e^{7x} \end{align*}.

    Now as for the nonhomogeneous solution, use \displaystyle \begin{align*} u = A\,x^2 + B\,x + C \end{align*}, which means \displaystyle \begin{align*} u' = 2A\,x + B  \end{align*} and \displaystyle \begin{align*} u'' = 2A \end{align*}. Substituting into the DE gives

    \displaystyle \begin{align*} u'' - 9u' + 14u &= x^2 \\ 2A - 9\left( 2A\,x + B \right) + 14 \left( A\, x^2 + B\, x + C \right) &= x^2 \\ 2A - 18A\,x - 9B + 14A\, x^2 + 14B\, x + 14C &= x^2 \\ 14A\, x^2 + \left( 14B - 18A \right) x + 2A - 9B + 14C &= 1x^2 + 0x + 0 \\ 14A = 1, 14B - 18A = 0, 2A - 9B + 14C &= 0 \\ A = \frac{1}{14}, B = \frac{9}{98}, C &= \frac{67}{1372} \end{align*}

    Therefore the nonhomogeneous solution is \displaystyle \begin{align*} \frac{1}{14}x^2 + \frac{9}{98}x + \frac{67}{1372} \end{align*}, so putting the homogeneous and nonhomogeneous solutions together gives the solution for \displaystyle \begin{align*} u \end{align*} as

    \displaystyle \begin{align*} u &= C_1\, e^{2x} + C_2\, e^{7x} + \frac{1}{14}x^2 + \frac{9}{98}x + \frac{67}{1372} \\ y' &= C_1 \, e^{2x} + C_2\, e^{7x} + \frac{1}{14}x^2 + \frac{9}{98}x + \frac{67}{1372} \\ y &= \frac{C_1}{2} \, e^{2x} + \frac{C_2}{7} \, e^{7x} + \frac{1}{42} x^3 + \frac{9}{196} x^2 + \frac{67}{1372}x + C_3  \end{align*}
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