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I'm getting $A = \frac{-3}{68}$ and $B = \frac{1}{34}$ for the guess $y_p = A\cos(2x)+B\sin(2x)$ using the method of undetermined coefficients. Apparently this answer is wrong (with $A$ and $B$ plugged in). Can anyone help me find the right one?

Thanks!

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    Since $\cos^2(x) = \frac{1}{2}\cos(2x) + \frac{1}{2}$, your particular solution $y_p$ better have a constant in it.2012-10-20

3 Answers 3

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Note that the right hand side is ${\cos ^2}(x)$, so want to choose something similar as our particular solution. But $\cos (2x) = {\cos ^2}(x) - {\sin ^2}(x) = 2{\cos ^2}(x) - 1$ so ${\cos ^2}(x) = \frac{1}{2} + \frac{1}{2}\cos (2x)$ and we can look for a particular solution of the form ${y_p} = A + B\sin (2x) + C\cos (2x).$

Then you need to use the initial conditions to determine the coefficients or plug ${y_p}$ into the differential equation, collect like terms, and equate the coefficients.

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Here is a method to determine the PI using the "operator method", if $f(-a^2)\ne0$, then $\frac{1}{f(D^2)}\cos(ax+b)=\frac{1}{f(-a^2)}\cos(ax+b)$ First note that $f(D)=5D^2+8D+8$

Now $\cos^2x=1/2+\cos2x/2$ and $1/f(D)(1/2)=(1/8)[1+D(1+5D/8)]^{-1}(1/2)=(1/8)(1/2)=1/16$

Also, $\frac{1}{f(D)}\cos2x=\frac{1}{5(-4)+8D+8}\cos2x=\frac{1}{4(2D-3)}\cos2x=\frac{2D+3}{4(4D^2-9)}\cos2x$

$=\frac{1}{4(-4.4-9)}[2D+3]\cos2x=-\frac{3\cos2x-4\sin2x}{100}$

Now add these two results to get $y_p$.

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See below for the details. Your equation is

$\left( {5{D^2} + 8D + 8} \right)y = {\cos ^2}\left( x \right)$

Make that into

$\left( {5{D^2} + 8D + 8} \right)y = \frac{{1 + \cos \left( {2x} \right)}}{2}$

and apply $D$

$\eqalign{ & \left( {5{D^2} + 8D + 8} \right)y = \frac{{1 + \cos \left( {2x} \right)}}{2} \cr & \left( {5{D^2} + 8D + 8} \right)y' = - \sin \left( {2x} \right) \cr & \left( { - 5{D^2} - 8D - 8} \right)y' = \sin \left( {2x} \right) \cr} $

Thus, we have $\eqalign{ & - 5{D^2} - 8D - 8 = \left( { - 5{D^2} - 8} \right) - 8D \cr & \varphi \left( D \right) = - 5D - 8 \cr & \xi \left( D \right) = - 8 \cr} $

and then

$y' = \frac{{3\sin 2x + 4\cos 2x}}{{100}}{\text{ }}$

and you solution is $y = \frac{{ - 3\cos 2x + 4\sin 2x}}{{200}} + C{ _1}$

I leave it to you to find $C_1$.


I know the following is not completely rigorous, but it has its justifications.

As a generalization of Tapu's solution, suppose we're given

$\phi(D)=\sum_{k=0}^m a_kD^k$

where $D=\dfrac d {dx}$. Since $D^2\sin(ax)=-a^2\sin (ax)$, we have that

$\phi(D^2)\sin(ax)=\phi(-a^2)\sin(ax)$

But we may write

$\phi (D) = \sum\limits_{k = 0}^{m'} {{a_{2k}}{D^{2k}}} + \sum\limits_{k = 0}^{m'} {{a_{2k + 1}}{D^{2k + 1}}} $

whence

$\phi (D) = \varphi \left( {{D^2}} \right) + D\xi \left( {{D^2}} \right)$

thus an equation of the form

$\phi (D)y = \sin ax$

becomes $\left[ {\varphi \left( {{D^2}} \right) + D\xi \left( {{D^2}} \right)} \right]y = \sin ax$

We apply the "conjugate" of the operator to the equation, to get $\eqalign{ & \left[ {{\varphi ^2}\left( {{D^2}} \right) - {D^2}{\xi ^2}\left( {{D^2}} \right)} \right]y = \left[ {\varphi \left( {{D^2}} \right) - D\xi \left( {{D^2}} \right)} \right]\sin ax \cr & \left[ {{\varphi ^2}\left( {{D^2}} \right) - {D^2}{\xi ^2}\left( {{D^2}} \right)} \right]y = \varphi \left( { - {a^2}} \right)\sin ax - a\xi \left( { - {a^2}} \right)\cos ax \cr & \left[ {{\varphi ^2}\left( {{D^2}} \right) - {D^2}{\xi ^2}\left( {{D^2}} \right)} \right]y = \alpha \sin ax + \beta \cos ax \cr & \Phi \left( {{D^2}} \right)y = \alpha \sin ax + \beta \cos ax \cr} $

Assuming $\Phi \left( { - {a^2}} \right) \ne 0$ we may write $\frac{{\sin ax}}{{\Phi \left( { - {a^2}} \right)}} = \frac{{\sin ax}}{{\Phi \left( {{D^2}} \right)}}$ so that

$\eqalign{ & y = \frac{{\alpha \sin ax + \beta \cos ax}}{{\Phi \left( { - {a^2}} \right)}} \cr & y = \frac{{\varphi \left( { - {a^2}} \right)\sin ax - a\xi \left( { - {a^2}} \right)\cos ax}}{{{\varphi ^2}\left( { - {a^2}} \right) + {a^2}{\xi ^2}\left( { - {a^2}} \right)}} \cr} $

For the cosine, you get $y = \frac{{\varphi \left( { - {a^2}} \right)\cos ax + a\xi \left( { - {a^2}} \right)\sin ax}}{{{\varphi ^2}\left( { - {a^2}} \right) + {a^2}{\xi ^2}\left( { - {a^2}} \right)}}$

As an example, consider

$(D^2-3D+2)y=\sin 3x$

Then $\phi(D)=D^2-3D+2=D^2+2-3D=\varphi(D^2)-D \xi(D^2)$ where $\varphi(D)=D+2$ and $\xi(D)=-3$. Thus $y = \frac{{ - 7\sin 3x - 3\left( { - 3} \right)\cos 3x}}{{49 + 9{{\left( { - 3} \right)}^2}}} = \frac{{9\cos 3x - 7\sin 3x}}{{130}}$

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    @glebovg Let the OP decide. I don't think it is necessary, but it isn't unnecessary either.2012-10-20