Anahi Solomon

2022-03-22

Which of the following is not a solution of $y{}^{″}+4y=0$
(A) $4\mathrm{cos}\left(2x\right)$
(B) $5\mathrm{sin}\left(2x\right)$
(C) $\mathrm{sin}\left(2x\right)\mathrm{cos}\left(2x\right)$
(D) $4\mathrm{cos}\left(2x\right)+5\mathrm{sin}\left(2x\right)$

undodaonePvopxl24

The given equation is a second order linear non homogeneous differential equation with constant coefficient.
The general solution for $a\left(x\right){y}^{″}+b\left(x\right){y}^{\prime }+c\left(x\right)y=g\left(x\right)$
The general solution of the given differential equation can be written as
$y={y}_{h}+{y}_{p}$
${y}_{h}$ is the solution to the homogeneous ODE $a\left(x\right){y}^{″}+b\left(x\right){y}^{\prime }+c\left(x\right)y=0$
Here the given ODE is the homogeneous equation.
The complementary solution for the given equation is:
$y{}^{″}+4y=0$
Rewrite the equation with $y={e}^{\gamma \cdot x}$
$\left({e}^{\gamma \cdot x}\right){}^{″}+4{e}^{\gamma \cdot x}=0$
${e}^{\gamma \cdot x}\left({\gamma }^{2}+4\right)=0$

For two complex roots ${\gamma }_{1}\ne q{\gamma }_{2}$, where ${\gamma }_{1}=a+i\cdot b$, ${\gamma }_{2}=a-i\cdot b$
the general solution takes the form: $y={e}^{ax}\left({c}_{1}\mathrm{cos}\left(bx\right)+{c}_{2}\mathrm{sin}\left(bx\right)\right)$
$y\left(x\right)={e}^{0}\left({c}_{1}\mathrm{cos}\left(2x\right)+{c}_{2}\mathrm{sin}\left(2x\right)\right)$
$y={c}_{1}\mathrm{cos}\left(2x\right)+{c}_{2}\mathrm{sin}\left(2x\right)$
Now, there are four options and one option is incorrect:
$y={c}_{1}\mathrm{cos}\left(2x\right)+{c}_{2}\mathrm{sin}\left(2x\right)$
(A) $4\mathrm{cos}\left(2x\right)$ is possible when ${c}_{1}=4$ and ${c}_{2}=0$
(B) $5\mathrm{sin}\left(2x\right)$ is possible when ${c}_{1}=0$ and ${c}_{2}=5$
(C) $\mathrm{sin}2x\cdot \mathrm{cos}\left(2x\right)$ is not possible
(D) $4\mathrm{cos}\left(2x\right)+5\mathrm{sin}\left(2x\right)$ is possible when ${c}_{1}=4$ and ${c}_{2}=5$
Hence the solution (C) is not possible

Jeffrey Jordon