solve by variation of parameters

**kithy** · June 30th 2008, 04:02 PM

need urgent help solving this DE by variation of parameters

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

any help is greatly appreciated

thank you

**TheEmptySet** · June 30th 2008, 06:07 PM

Originally Posted by kithy

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

$W=\begin{vmatrix} \sin(t) && \cos(t) \\ \cos(t) && -\sin(t) \\ \end{vmatrix}=-1$

$W_1=\begin{vmatrix} 0 && \cos(t) \\ \sec(t)\tan(t) && -\sin(t) \\ \end{vmatrix}=-\tan(t)$

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

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

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

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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