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Math Help - [SOLVED] initial value problem

  1. #1
    Member ronaldo_07's Avatar
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    [SOLVED] initial value problem

    Solve the following initial value problem
    y′′ + 9y = 3 cos(3x), y(0) = 0, y(0) = 0 and write the final solution using real valued functions. Sketch the graph of the solution in a x-y diagram.

    Can someone help me solve this?
    This is what i done so far

    y′′ + 9y - 3 cos(3x)= 0
    z^2 +9z -3 cos(3x)= 0

    z^2+6z+3z -3cos(3x)=0
    z(z+6) + 3(z-cos(3x))=0

    (z+3)(z+6)(z-cos(3x))=0

    Now im really stuck on what to do. Please help. Thanks.

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  2. #2
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    Quote Originally Posted by ronaldo_07 View Post
    Solve the following initial value problem
    y′′ + 9y = 3 cos(3x), y(0) = 0, y(0) = 0 and write the final solution using real valued functions. Sketch the graph of the solution in a x-y diagram.

    Can someone help me solve this?
    This is what i done so far

    y′′ + 9y - 3 cos(3x)= 0
    z^2 +9z -3 cos(3x)= 0

    z^2+6z+3z -3cos(3x)=0
    z(z+6) + 3(z-cos(3x))=0

    (z+3)(z+6)(z-cos(3x))=0

    Now im really stuck on what to do. Please help. Thanks.

    Nothing that you have done so far makes sense. Why did you replace y with z and y" with z^2? If you are trying to find the "characteristic equation" for this differential equation, that only works for homogeneous equations. First solve the associated homogeneous equation.
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by ronaldo_07 View Post
    Solve the following initial value problem




    y′′ + 9y = 3 cos(3x), y(0) = 0, y(0) = 0 and write the final solution using real valued functions. Sketch the graph of the solution in a x-y diagram.


    Can someone help me solve this?
    This is what i done so far

    y′′ + 9y - 3 cos(3x)= 0
    z^2 +9z -3 cos(3x)= 0

    z^2+6z+3z -3cos(3x)=0
    z(z+6) + 3(z-cos(3x))=0

    (z+3)(z+6)(z-cos(3x))=0

    Now im really stuck on what to do. Please help. Thanks.
    You need to break this problem up into two parts.

    Fist we need to solve the homogenious equation

    y''+9y=0 \implies m^2+9m=0 \implies m=\pm 3i

    So this gives us the complimentry solution

    y_c=c_1\cos(3x)+c_2\sin(3x)

    Now we need to find a particualr solution. There are a few different ways to do this, but here is one way.

    w=\begin{vmatrix} <br />
\cos(3x) & \sin(3x) \\<br />
-3\sin(3x) & 3\cos(3x) <br />
\end{vmatrix}=3

    w_1=\begin{vmatrix} <br />
0 & \sin(3x) \\<br />
3\cos(3x) & 3\cos(3x) <br />
\end{vmatrix}=-\sin(3x)\cos(3x)

    w_2=\begin{vmatrix} <br />
\cos(3x) & 0 \\<br />
-3\sin(3x) & 3\cos(3x) <br />
\end{vmatrix}=\cos^2(3x)

     <br />
u_1'=\frac{w_1}{w} \implies u=-\int \sin(3x)\cos(3x)dx=\frac {1}{6} \cos^2(3x) <br />

    u_2'=\frac{w_2}{w} \implies \int \cos^2(3x)dx=\frac{1}{2}x+\frac{1}{6}\sin(3x)\cos(  3x)

    so

    y_p=\frac{1}{6}\cos^3(3x)+(\frac{1}{2}x+\frac{1}{6  }\sin(3x)\cos(3x))\sin(3x)=

    y_p=\frac{1}{6}\cos^3(3x)+\frac{x}{2}\sin(3x)+\fra  c{1}{6}\sin^2(3x)\cos(3x)=

    y_p=\frac{1}{6}\cos^3(3x)+\frac{x}{2}\sin(3x)+\fra  c{1}{6}(1-\cos^2(3x))\cos(3x)=

    y_p=\frac{x}{2}\sin(3x)+\frac{1}{6}\cos(3x)

    so the final solution is

    y=y_c+y_p=c_1\cos(3x)+c_2\sin(3x)+\frac{x}{2}\sin(  3x)+\frac{1}{6}\cos(3x)

    Now using the intial conditions we get

    0=c_1+\frac{1}{6} \iff c_1=-\frac{1}{6}

    y=c_2\sin(3x) + \frac{x}{2}\sin(3x)

    taking the derivative we get

    y'=3c_2\cos(3x)+\frac{1}{2}\sin(3x)+\frac{3x}{2}\c  os(3x)

    using the 2nd ic we get

    0=3c_2 \iff c_2=0

    y=\frac{x}{2}\sin(3x)
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