use trigonometric identities to find general solutions of the equation. y + 81y = sin^4x. the general…

Explanation:

Step1: Find the complementary - function

The homogeneous equation is $y'' + 81y=0$. The characteristic equation is $r^{2}+81 = 0$. Solving for $r$, we have $r^{2}=-81$, so $r=\pm9i$. The complementary - function $y_c(x)=C_1\cos(9x)+C_2\sin(9x)$.

Step2: Rewrite $\sin^{4}x$ using trigonometric identities

We know that $\sin^{2}x=\frac{1 - \cos(2x)}{2}$. Then $\sin^{4}x=(\sin^{2}x)^{2}=\left(\frac{1 - \cos(2x)}{2}\right)^{2}=\frac{1-2\cos(2x)+\cos^{2}(2x)}{4}$. And since $\cos^{2}(2x)=\frac{1 + \cos(4x)}{2}$, we get $\sin^{4}x=\frac{1-2\cos(2x)+\frac{1 + \cos(4x)}{2}}{4}=\frac{2-4\cos(2x)+1+\cos(4x)}{8}=\frac{3}{8}-\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$.

Step3: Find the particular - solution

We assume a particular solution of the form $y_p(x)=A + B\cos(2x)+C\sin(2x)+D\cos(4x)+E\sin(4x)$. Taking the first - derivative: $y_p'(x)=-2B\sin(2x)+2C\cos(2x)-4D\sin(4x)+4E\cos(4x)$. Taking the second - derivative: $y_p''(x)=-4B\cos(2x)-4C\sin(2x)-16D\cos(4x)-16E\sin(4x)$. Substituting $y_p$, $y_p''$ into the non - homogeneous equation $y'' + 81y=\sin^{4}x$: [ \begin{align*} (-4B\cos(2x)-4C\sin(2x)-16D\cos(4x)-16E\sin(4x))+81(A + B\cos(2x)+C\sin(2x)+D\cos(4x)+E\sin(4x))&=\frac{3}{8}-\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)\ 81A+(81B - 4B)\cos(2x)+(81C - 4C)\sin(2x)+(81D-16D)\cos(4x)+(81E - 16E)\sin(4x)&=\frac{3}{8}-\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x) \end{align*} ] For the constant term: $81A=\frac{3}{8}$, so $A=\frac{1}{216}$. For the $\cos(2x)$ term: $77B=-\frac{1}{2}$, so $B =-\frac{1}{154}$. For the $\sin(2x)$ term: $77C = 0$, so $C = 0$. For the $\cos(4x)$ term: $65D=\frac{1}{8}$, so $D=\frac{1}{520}$. For the $\sin(4x)$ term: $65E = 0$, so $E = 0$. So $y_p(x)=\frac{1}{216}-\frac{1}{154}\cos(2x)+\frac{1}{520}\cos(4x)$.

Step4: Find the general solution

The general solution of the non - homogeneous equation is $y(x)=y_c(x)+y_p(x)=C_1\cos(9x)+C_2\sin(9x)+\frac{1}{216}-\frac{1}{154}\cos(2x)+\frac{1}{520}\cos(4x)$.

Answer:

$y(x)=C_1\cos(9x)+C_2\sin(9x)+\frac{1}{216}-\frac{1}{154}\cos(2x)+\frac{1}{520}\cos(4x)$