The given equation is $$y^{(4)}+5y^{''}+4y= \sin(x)+\cos(2x)$$Using the auxiliary equation to find the roots result with $m_{1,2}=\pm i$ and $m_{3,4}=\pm 2i$. Usually the equation characteristic is $y= C_1e^{M_1}+C_2e^{M_2}$, but because we have imaginary terms we change the positive $i$ values into $\cos(x)$ and negatives into $\sin(x)$.
Meaning the characteristic function should look like
$$y_c=C_1\sin(x)+C_2 \cos(x)+C_3 \sin(2x)+C_4\cos(2x)$$And knowing that $g(x)= \sin(x)+\cos(2x)$ we can see that there are duplicate within the two functions. So we must multiply $g(x)$ by $x$ in order to solve. After multiplying out the $x$ our $y_p$ becomes$$y_p=Ax\sin(x) +Bx \cos(x)+Cx\sin(2x)+dx\cos(2x)$$Now taking the fourth derivative gives us $$ y^{(4)}_p=Ax\sin(x)-4A\cos(x)+Bx\cos(x)+4B\sin(x)+16Cx\sin(2x) -20C\cos(2x)+16Dx\cos(2x)+20D\sin(2x)$$Now I plugged in each derivative into the given equation. After distributing and factoring my final answer is $$-6(A\cos(x)+B\sin(x))=\sin(x)+\cos(2x)$$How can I find the general solution from here?
My work is below if you want to check my arithmetic.!
2 Answers
$\begingroup$HINT
Start by noticing that
\begin{align*} y'''' - 5y'' + 4y = \sin(x) + \cos(2x) & \Longleftrightarrow (y'' - 4y)'' - (y'' - 4y) = \sin(x) + \cos(2x)\\\\ & \Longleftrightarrow z'' - z = \sin(x) + \cos(2x)\\\\ & \Longleftrightarrow (z' - z)' + (z' - z) = \sin(x) + \cos(2x)\\\\ & \Longleftrightarrow w' + w = \sin(x) + \cos(2x) \end{align*}
Can you take it from here?
$\endgroup$ $\begingroup$$$y^{(4)}-5y^{''}+4y= \sin(x)+\cos(2x)$$The characteristic polynomial is:$$r^4-5r^2+4=0$$$$r^2(r^2-4)-(r^2-4)=0$$$$(r^2-4)(r^2-1)=0$$$$(r-2)(r+2)(r-1)(r+1)=0$$$$\implies S_r=\{-1,1,2,-2\}$$The complementary solution is$$y_h=c_1e^x+c_2e^{-x}+c_3e^{2x}+c_4e^{-2x}$$The particular solution should be of the form :$$y_p=A\sin x + B \cos(2x)$$
$\endgroup$ 8
