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Thread: Nonhomogenous Linear Equation

  1. #1
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    Nonhomogenous Linear Equation

    Solve the following differential equation:

    $\displaystyle y^{''}+y=e^x+x^3$

    so starting off with the auxiliary equation I have:

    $\displaystyle r^2+1=0$ where the only available solution is $\displaystyle \pm i$

    so I would have:

    $\displaystyle y_c=c_1 \cos(x)+c_2 \sin(x)$

    working on the particular solution I let $\displaystyle y_{p1}(x) = Ae^x$

    substituting for $\displaystyle y_{p1}$ I come up $\displaystyle Ae^x+Ae^x=e^x \Rightarrow 2Ae^x=e^x \Rightarrow A=\frac{1}{2} e^x$

    (this is the part that I get stuck on) Since $\displaystyle x^3$ is a polynomial of third degree I would have:

    $\displaystyle y_{p2}=Bx^3+Cx^2+Dx+E$

    where $\displaystyle y_{p2}^{''} = 6Bx+C$

    substituting I get:

    $\displaystyle (6Bx+C)+ (Bx^3+Cx^2+Dx+E) =x^3$, then

    $\displaystyle (6B+C)x^3+ (Bx^3+Cx^2+Dx+E) =x^3$

    so $\displaystyle (6B+C)=1$ and $\displaystyle (B+C+D+E)=0$ where I don't think that I'm approaching it correctly since i have 4 unknowns.

    any help would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by lllll View Post
    Solve the following differential equation:

    $\displaystyle y^{''}+y=e^x+x^3$

    so starting off with the auxiliary equation I have:

    $\displaystyle r^2+1=0$ where the only available solution is $\displaystyle \pm i$

    so I would have:

    $\displaystyle y_c=c_1 \cos(x)+c_2 \sin(x)$

    working on the particular solution I let $\displaystyle y_{p1}(x) = Ae^x$

    substituting for $\displaystyle y_{p1}$ I come up $\displaystyle Ae^x+Ae^x=e^x \Rightarrow 2Ae^x=e^x \Rightarrow A=\frac{1}{2} e^x$ Mr F says: $\displaystyle {\color{red} \Rightarrow A = \frac{1}{2}}$. Therefore $\displaystyle {\color{red}y_{p1}(x) = \frac{1}{2} e^x}$.

    (this is the part that I get stuck on) Since $\displaystyle x^3$ is a polynomial of third degree I would have:

    $\displaystyle y_{p2}=Bx^3+Cx^2+Dx+E$

    where $\displaystyle y_{p2}^{''} = 6Bx+C$ Mr F says: $\displaystyle {\color{red}y_{p2}^{''} = 6Bx+ {\color{blue}2} C}$. Note the blue bit. This will effect all the calculations that follow.

    substituting I get:

    $\displaystyle (6Bx+C)+ (Bx^3+Cx^2+Dx+E) =x^3$, then

    $\displaystyle (6B+C)x^3+ (Bx^3+Cx^2+Dx+E) =x^3$

    so $\displaystyle (6B+C)=1$ and $\displaystyle (B+C+D+E)=0$ where I don't think that I'm approaching it correctly since i have 4 unknowns.

    any help would be greatly appreciated.
    Some corrections in red. The bit in blue is important.
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    given the corrections, then would the following be correct:

    $\displaystyle (6Bx+2C)+ (Bx^3+Cx^2+Dx+E) =x^3$, then

    $\displaystyle (6B+2C)x^3+ (B+C+D+E) =x^3$, where

    $\displaystyle {\color{blue}(1)} \ (6B+2C) = 1 \ \mbox{and} \ {\color{blue}(2)} \ (B+C+D+E) = 0$

    $\displaystyle B= \frac{1-2C}{6}$ and substituting into the second equation

    which ends up giving me $\displaystyle \frac{2}{3} C+D+E = -\frac{1}{6} $

    if I let $\displaystyle D$ and $\displaystyle E$ equal to 0 then I get a solution for $\displaystyle C$, but I don't think that I can do that.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by lllll View Post
    Solve the following differential equation:

    $\displaystyle y^{''}+y=e^x+x^3$

    so starting off with the auxiliary equation I have:

    $\displaystyle r^2+1=0$ where the only available solution is $\displaystyle \pm i$

    so I would have:

    $\displaystyle y_c=c_1 \cos(x)+c_2 \sin(x)$

    working on the particular solution I let $\displaystyle y_{p1}(x) = Ae^x$

    substituting for $\displaystyle y_{p1}$ I come up $\displaystyle Ae^x+Ae^x=e^x \Rightarrow 2Ae^x=e^x \Rightarrow A=\frac{1}{2} e^x$

    (this is the part that I get stuck on) Since $\displaystyle x^3$ is a polynomial of third degree I would have:

    $\displaystyle y_{p2}=Bx^3+Cx^2+Dx+E$

    where $\displaystyle y_{p2}^{''} = 6Bx+C$

    substituting I get:

    $\displaystyle (6Bx+C)+ (Bx^3+Cx^2+Dx+E) =x^3$, then

    $\displaystyle (6B+C)x^3+ (Bx^3+Cx^2+Dx+E) =x^3$

    so $\displaystyle (6B+C)=1$ and $\displaystyle (B+C+D+E)=0$ where I don't think that I'm approaching it correctly since i have 4 unknowns.

    any help would be greatly appreciated.
    When dealing with the non-homogeneous equation, I'd try to approach this using the annihilator technique (for it gives us the exact form of our particular solution in case we can't make a good guess at it). You can read more about it here.

    Since $\displaystyle y^{\prime\prime}+y=e^x+x^3\implies (D^2+1)y=e^x+x^3$, the annihilator for $\displaystyle e^x=D-1$ and the annihilator for $\displaystyle x^3=D^4$. Thus, the Annihilator for $\displaystyle e^x+x^3=D^4(D-1)$.

    Therefore, $\displaystyle (D^2+1)y=e^x+x^3\implies D^4(D-1)(D^2+1)y=D^4(D-1)\left[e^x+x^3\right]$ $\displaystyle \implies D^4(D-1)(D^2+1)y=0$

    Then converting this over to the auxiliary equation, we have $\displaystyle r^4(r-1)(r^2+1)=0\implies r=0~\text{multiplicity 4}, r=1, r=\pm i$

    Therefore, $\displaystyle y_p=A+Bx+Cx^2+Dx^3+Ee^x$

    Thus, $\displaystyle y_p^{\prime\prime}=2C+6Dx+Ee^x$.

    Thus, $\displaystyle y_p^{\prime\prime}+y_p=e^x + x^3\implies A+2C+(B+6D)x+Cx^2+Dx^3+2Ee^x=x^3+e^x$.

    Therefore,

    $\displaystyle \begin{array}{rcrcrcrcrcr} A&~&~&+&2C&~&~&~&~&=&0\\ ~ &~&B&~&~&+&6D&~&~&=&0\\~&~&~&~&C&~&~&~&~&=&0\\~&~& ~&~&~&~&D&~&~&=&1\\~&~&~&~&~&~&~&~&2E&=&1\end{arra y}$

    Thus, $\displaystyle A=0,~B=-6,~C=0,~D=1,~E=\tfrac{1}{2}$

    Therefore, the solution to your DE is $\displaystyle y=c_1\cos x+c_2\sin x-6x+x^3+\tfrac{1}{2}e^x$

    Does this help?
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  5. #5
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    Quote Originally Posted by lllll View Post
    given the corrections, then would the following be correct:

    $\displaystyle (6Bx+2C)+ (Bx^3+Cx^2+Dx+E) =x^3$, then

    $\displaystyle (6B+2C)x^3+ (B+C+D+E) =x^3$, where Mr F says: I'm at a loss to understand how you could get this. Where have the x^2 terms and x terms gone on the left hand side? Where has the x^3 term come from?

    $\displaystyle {\color{blue}(1)} \ (6B+2C) = 1 \ \mbox{and} \ {\color{blue}(2)} \ (B+C+D+E) = 0$

    $\displaystyle B= \frac{1-2C}{6}$ and substituting into the second equation

    which ends up giving me $\displaystyle \frac{2}{3} C+D+E = -\frac{1}{6} $

    if I let $\displaystyle D$ and $\displaystyle E$ equal to 0 then I get a solution for $\displaystyle C$, but I don't think that I can do that.
    Re-arrange:

    $\displaystyle Bx^3 + Cx^2 + (6B + D)x + (2C + E) = x^3$.

    Equate coefficients:

    B = 1 .... (1)

    C = 0 .... (2)

    6B + D = 0 .... (3)

    2C + E = 0 .... (4)

    Solve simultaneously.
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