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My problem is the following: $y'' + 4y = 4\sin(2x)$

I tried making a particular solution to this by assuming that $y = A\sin(2x) + B\cos(2x)$

This gives me the derivate $y' = 2A\cos(2x) - 2B\sin(2x)$ and

The second derivate $y'' = -4A\sin(2x) - 4B\cos(2x)$

If I plug the second derivate and 4 times the equation into the starting problem:

I get: $-4A\sin(2x) - 4B\cos(2x) + 4[A\sin(2x) + B\cos(2x)] = 4\sin(2x) + 0cos(2x)$

As you can see, I get $0 = 4\sin(2x)$, which is impossible because $0$ is not $4$.

I can't find this problem anywhere, how do I go about solving this problem?

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    $\begingroup$ Hint: You have to choose $y_p = x( a \sin(2x) + b \cos(2x))$ because the homogeneous solution already contains $\sin(2x)$. See tutorial.math.lamar.edu/Classes/DE/…. $\endgroup$ – Moo Mar 4 '17 at 18:47
  • $\begingroup$ the solution is given by $$y \left( x \right) =\sin \left( 2\,x \right) {\it \_C2}+\cos \left( 2 \,x \right) {\it \_C1}-\cos \left( 2\,x \right) x $$ $\endgroup$ – Dr. Sonnhard Graubner Mar 4 '17 at 19:01
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I assume you first found your homogeneous solution.

$$y''+4y=0 \quad\implies\quad y_h = A\sin 2x + B\cos 2x$$

Your non-homogeneous equation has a forcing term proportional to $\sin 2x$.

$$y''+4y=4\sin 2x$$

You know that guessing a particular solution of the form $y_p=y_h$ makes the LHS zero, which isn't helpful. But you need the $\sin2x$ to appear on the left. The next best thing would be to incorporate a polynomial factor and hope it gets derived away. i.e., your particular solution is anticipated to be

$$y_p = xy_h = x A\sin2x + x B\cos2x$$

Then,

\begin{align} y_p' &= y_h +2x A\cos2x - 2x B\sin2x \\ \\ y_p'' &= y_h' +2 A\cos2x - 2 B\sin2x -4x A\sin2x - 4x B\cos2x \\ &= 4A\cos2x- 4B\sin2x -4x A\sin2x - 4x B\cos2x \\ &= 4(A\cos2x- B\sin2x) -4x y_h \end{align}

Upon substituting the particular solution back into the differential equation, \begin{align} 4(A\cos2x- B\sin2x) -4xy_h + 4xy_h = 4\sin 2x \end{align}

we see that $A=0,B=-1$ in the particular solution. Thus $y_p = -x\cos2x$ and

$$y = c_1\sin2x+c_2\cos2x -x\cos2x$$

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  • $\begingroup$ Hey, thanks for the answer, but there is one thing I don't understand. First, you write yp = xyh = xAsin2x+xBcos2x with equals in between the three elements. Then in the next row after a derivation, you add a yh. Why? $\endgroup$ – Slayahh Mar 7 '17 at 10:05
  • $\begingroup$ I noticed the particular solution $y_p = x A\sin2x + x B\cos2x$ was of the form of the homogeneous solutions times $x$, so I started using $y_p=xy_h$ in some places, mainly to tidy up the work. I edited in a few lines to hopefully clarify: Note that $y_p' = y_h + xy_h'$. Effectively, every time you see $y_h$ in the bottom half of the work, substitute it out mentally for $ A\sin2x + B\cos2x$. $\endgroup$ – zahbaz Mar 7 '17 at 16:42

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