Question #c6898

Dec 10, 2017

$y = {c}_{1} \cdot \sin 2 x + {c}_{2} \cdot \cos 2 x + \frac{1}{4} x \sin 2 x - \frac{1}{12} \cos 4 x$

Explanation:

$\left({D}^{2} + 4\right) y = \cos 2 x + \cos 4 x$

Characteristic equation of differential equation ${r}^{2} + 4 = 0$

Roots of it are ${r}_{1} = - 2 i$ and ${r}_{2} = 2 i$

Hence homogeneous part of it ${y}_{h} = {c}_{1} \cdot \sin 2 x + {c}_{2} \cdot \cos 2 x$

Due to $\cos 2 x$ is root of homogeneous part of it, particular solution of it is ${y}_{p} = A x \sin 2 x + B x \cos 2 x + C \sin 4 x + D \cos 4 x$

Consequently,

${D}^{2} \left(A x \sin 2 x + B x \cos 2 x + C \sin 4 x + D \cos 4 x\right) + 4 \left(A x \sin 2 x + B x \cos 2 x + C \sin 4 x + D \cos 4 x\right) = \cos 2 x + \cos 4 x$

$4 A \cos 2 x - 4 A x \sin 2 x - 4 B \sin 2 x - 4 B x \cos 2 x - 16 C \sin 4 x - 16 D \cos 4 x + 4 A x \sin 2 x + 4 B x \cos 2 x + 4 C \sin 4 x + 4 D \cos 4 x = \cos 2 x + \cos 4 x$

$4 A \cos 2 x - 4 B \sin 2 x - 12 C \sin 4 x - 12 D \cos 4 x = \cos 2 x + \cos 4 x$

After equating coefficients,

$4 A = 1$, $- 4 B = 0$, $- 12 C = 0$ and $- 12 D = 1$

So, $B = C = 0$, $A = \frac{1}{4}$ and $D = - \frac{1}{12}$

Consequently, ${y}_{p} = \frac{1}{4} x \sin 2 x - \frac{1}{12} \cos 4 x$

Thus, $y = {y}_{h} + {y}_{p} = {c}_{1} \cdot \sin 2 x + {c}_{2} \cdot \cos 2 x + \frac{1}{4} x \sin 2 x - \frac{1}{12} \cos 4 x$