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Math Help - solve by variation of parameters

  1. #1
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    solve by variation of parameters

    need urgent help solving this DE by variation of parameters

    y'' + y = sec (theta) * tan (theta)

    any help is greatly appreciated

    thank you
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by kithy View Post
    need urgent help solving this DE by variation of parameters

    y'' + y = sec (theta) * tan (theta)

    any help is greatly appreciated

    thank you
    Hello kithy,

    First we need to find the complimentry solution to the associated homogenious equation.

    y_c=c_1\sin(t)+c_2\cos(t)

    Now we know that our particular solution will be of the form.

    y_p=u_1(t)\sin(t)+u_2(t)\cos(t)

    So the wronskian is

    W=\begin{vmatrix} <br />
\sin(t) && \cos(t) \\<br />
\cos(t) && -\sin(t) \\<br />
\end{vmatrix}=-1


    W_1=\begin{vmatrix} <br />
0 && \cos(t) \\<br />
\sec(t)\tan(t) && -\sin(t) \\<br />
\end{vmatrix}=-\tan(t)


    W_2=\begin{vmatrix} <br />
\sin(t) && 0 \\<br />
\cos(t) && \sec(t)\tan(t) \\<br />
\end{vmatrix}=\tan^2(t)=\sec^2(t)-1

    u_1(t)=\int \frac{W_1}{W}dt=\int \tan(t)dt=-\ln|\cos(t)|

    u_2(t)=\int \frac{W_2}{W}dt=\int -\sec^2(t)+1 dt =t-\tan(t)

    So finally our solution is

    y=c_1\sin(t)+c_2\cos(t)+(-\ln|\cos(t)|)\sin(t)+(t-\tan(t))\cos(t)

    simplifying gives

    y=c_1\sin(t)+c_2\cos(t)-\sin(t)\ln|\cos(t)|+t\cos(t)-\sin(t)
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