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Thread: 2nd Order differential equations

  1. #1
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    Smile 2nd Order differential equations

    Need help to complete this question,

    $\displaystyle \frac{d^2 y}{dx^2}-5\frac{dy}{dx}+6y = x^2$

    $\displaystyle D^2-5D+6 = 0$

    $\displaystyle D^2-3D-2D+6 = 0$

    $\displaystyle D(D-3) -2(D-3) = 0$

    $\displaystyle (D-2)(D-3) = 0$

    $\displaystyle D = 2,3$

    $\displaystyle \therefore y_c = Ae^{2x}+Be^{3x}$

    $\displaystyle (D^2-5D+6)y_p = x^2$

    $\displaystyle y_p = \frac{x^2}{(D-2)(D-3)}$

    ...................................
    ...................................
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  2. #2
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    Quote Originally Posted by ashura View Post
    Need help to complete this question,

    $\displaystyle \frac{d^2 y}{dx^2}-5\frac{dy}{dx}+6y = x^2$

    $\displaystyle D^2-5D+6 = 0$

    $\displaystyle D^2-3D-2D+6 = 0$

    $\displaystyle D(D-3) -2(D-3) = 0$

    $\displaystyle (D-2)(D-3) = 0$

    $\displaystyle D = 2,3$

    $\displaystyle \therefore y_c = Ae^{2x}+Be^{3x}$

    $\displaystyle (D^2-5D+6)y_p = x^2$

    $\displaystyle y_p = \frac{x^2}{(D-2)(D-3)}$

    ...................................
    ...................................
    I have seen that operator before but not sure how it works.
    Just look for a solution of the form $\displaystyle Ax^2+Bx+C$ by undetermined coefficients.

    Thus,
    $\displaystyle 6y=6Ax^2+6Bx+6C$
    $\displaystyle -5y'=-10Ax-5B$
    $\displaystyle y''=2A$
    Thus, adding
    $\displaystyle 6Ax^2+(6B-10A)x+(6C-5B+2A)=x^2$
    So solve,
    $\displaystyle 6A=1$
    $\displaystyle 6B-10A=0$
    $\displaystyle 6C-5B+2A=0$

    And your general solution is correct!
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  3. #3
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    Hello, ashura!

    $\displaystyle \frac{d^2 y}{dx^2}-5\frac{dy}{dx}+6y \:= \:x^2$

    $\displaystyle D^2-5D+6 \:= \:0\quad\Rightarrow\quad (D-2)(D-3) \:= \:0\quad\Rightarrow\quad D \,= \,2,\,3$

    . . $\displaystyle \therefore \;y_c \:= \:Ae^{2x}+Be^{3x}$


    $\displaystyle (D^2-5D+6)y_p \:= \:x^2\quad\Rightarrow\quad y_p \:= \:\frac{x^2}{(D-2)(D-3)}$

    Evidently, you're using the Method of Operators.

    You're expected to know that: .$\displaystyle \boxed{\frac{1}{D - a}\!\cdot\!f(x) \:=\:e^{ax}\int e^{-ax}f(x)\,dx}$


    Hence: .$\displaystyle \frac{1}{D-3}(x^2)\;=\;e^{3x}\underbrace{\int e^{-3x}x^2\,dx}_{\text{integrate by parts}}$

    . . $\displaystyle = \;e^{3x}\left[-\frac{1}{3}x^2e^{-3x} - \frac{2}{9}xe^{-3x} - \frac{2}{27}e^{-3x} + C_1\right] $

    . . $\displaystyle = \;-\frac{1}{3}x^2 - \frac{2}{9}x - \frac{2}{27} + C_1e^{3x} \;= \;-\frac{1}{27}\left(9x^2 + 6x + 2 + C_1e^{3x}\right) $

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    I'll let you take it through the second stage . . .

    $\displaystyle \frac{1}{D - 2}\left[-\frac{1}{27}\left(9x^2 + 6x + 2 + C_1e^{3x}\right)\right]$

    . . . $\displaystyle = \;e^{2x}\int e^{-2x}\left[-\frac{1}{27}\left(9x^2 + 6x + 2 + C_1e^{3x}\right)\right]\,dx $

    . . . $\displaystyle = \;-\frac{1}{27}e^{2x}\int \left[\left(9x^2 + 6x + 2\right)e^{-2x} + C_1e^x\right]\,dx$ . . . etc.

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  4. #4
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    Wink

    Thanks for your help. I can now complete the solution.
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