Results 1 to 2 of 2

Thread: Difficult DiffyQ

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    124

    Difficult DiffyQ

    y'' - 4y = xe^(x)

    Use the method of vartiation of parameters to find the general solution of the given differential equation.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5
    Quote Originally Posted by bearej50 View Post
    y'' - 4y = xe^(x)

    Use the method of vartiation of parameters to find the general solution of the given differential equation.
    First solve the homogeneous equation $\displaystyle y^{\prime\prime}-4y=0$.

    The characteristic equation is $\displaystyle r^2-4=0\implies r=\pm 2$.

    Thus, the complimentary solution is $\displaystyle y=c_1e^{-2x}+c_2e^{2x}$.

    Now, let $\displaystyle y_1\!\left(x\right)=e^{-2x}$ and $\displaystyle y_2\!\left(x\right)=e^{2x}$.

    We now need to find $\displaystyle u_1^{\prime}\!\left(x\right)$ and $\displaystyle u_2^{\prime}\!\left(x\right)$

    $\displaystyle u_1^{\prime}\!\left(x\right)=\frac{\displaystyle\b egin{vmatrix}0 & y_2\\f\!\left(x\right)&y_2^{\prime}\end{vmatrix}}{ \displaystyle\begin{vmatrix}y_1 & y_2\\y_1^{\prime} & y_2^{\prime}\end{vmatrix}}=\frac{\displaystyle\beg in{vmatrix}0 & e^{2x}\\xe^{x} & 2e^{2x}\end{vmatrix}}{\displaystyle\begin{vmatrix} e^{-2x} & e^{2x}\\-2e^{-2x} & 2e^{2x}\end{vmatrix}}=\tfrac{1}{4}xe^{3x}$

    Thus, $\displaystyle u_1\!\left(x\right)=\tfrac{1}{4}\int xe^{3x}\,dx$

    I leave it for you to find $\displaystyle u_1\!\left(x\right)$ (See spoiler to check your answer).

    Spoiler:
    $\displaystyle u_1\!\left(x\right)=\tfrac{1}{36}e^{3x}\left(3x-1\right)$


    $\displaystyle u_2^{\prime}\!\left(x\right)=\frac{\displaystyle\b egin{vmatrix}y_1 & 0\\ y_1^{\prime} & f\!\left(x\right)\end{vmatrix}}{\displaystyle\begi n{vmatrix}y_1 & y_2\\y_1^{\prime} & y_2^{\prime}\end{vmatrix}}=\frac{\displaystyle\beg in{vmatrix}e^{-2x} &0\\-2e^{-2x} & xe^{x}\end{vmatrix}}{\displaystyle\begin{vmatrix}e ^{-2x} & e^{2x}\\-2e^{-2x} & 2e^{2x}\end{vmatrix}}=\tfrac{1}{4}xe^{-x}$

    Thus, $\displaystyle u_2\!\left(x\right)=\tfrac{1}{4}\int xe^{-x}\,dx$

    I leave it for you to find $\displaystyle u_2\!\left(x\right)$ (See spoiler to check your answer).

    Spoiler:
    $\displaystyle u_2\!\left(x\right)=-\tfrac{1}{4}e^{-x}\left(x+1\right)$


    Once you find $\displaystyle u_1\!\left(x\right)$ and $\displaystyle u_2\!\left(x\right)$, the solution to your differential equation would be $\displaystyle y=u_1\!\left(x\right)e^{-2x}+u_2\!\left(x\right)e^{2x}=\dots$ (See spoiler to check your answer)

    Spoiler:
    $\displaystyle y=\tfrac{1}{36}e^{x}\left(3x-1\right)-\tfrac{1}{4}e^{x}\left(x+1\right)=\boxed{-\tfrac{1}{18}e^x\left(3x+5\right)}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. help w/ difficult IVP
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Apr 3rd 2010, 05:44 PM
  2. difficult(for me) DE
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Apr 3rd 2010, 05:49 AM
  3. solving diffyQ for calc2
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Nov 25th 2009, 02:58 PM
  4. DiffyQ
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Feb 20th 2009, 06:54 AM
  5. not to difficult
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: Nov 18th 2006, 08:51 PM

Search Tags


/mathhelpforum @mathhelpforum