Answer to Question #284752 in Differential Equations for Omk

Question

(D ^ 2 + 3D + 2) * y = sin 2x + 5 degrees + log 3

Answer 1

The given question is;

$\displaystyle (D^2+3D+2)y=\sin(2x+5)+\log3$

The Auxiliary equation of the given DE in terms of variable m is;

$\displaystyle m^2+3m+2=0\\$

$\displaystyle \Rightarrow(m+1)(m+2)=0\\ \Rightarrow m=-1,-2\\$

Hence, the complementary function is;

$\displaystyle y_c=c_1e^{-x}+c_2e^{-2x}$, where $\displaystyle c_1$ and $\displaystyle c_2$ are arbitrary constants.

Next, to calculate the particular integral $\displaystyle (y_p)$ we proceed as follows;

$\displaystyle y_p=\frac{1}{D^2+3D+2}\sin2\left(x+\frac{5}{2}\right)+\frac{1}{D^2+3D+2}(\log3)\\ \quad=\frac{1}{-4+3D+2}\sin2\left(x+\frac{5}{2}\right)+\log3\frac{1}{D^2+3D+2}e^{0x}\\ \quad=\frac{1}{3D-2}\sin2\left(x+\frac{5}{2}\right)+\frac{\log3}{2}\\ \quad=\frac{3D+2}{9D^2-4}\sin2\left(x+\frac{5}{2}\right)+\frac{\log3}{2}\\ \quad=\frac{3D+2}{-40}\sin2\left(x+\frac{5}{2}\right)+\frac{\log3}{2}\\ \quad=\frac{-1}{40}\left[3D\left(\sin(2x+5)\right)+2\sin(2x+5)\right]+\frac{\log3}{2}\\ \quad=\frac{-1}{40}[6\cos(2x+5)+2\sin(2x+5)]+\frac{\log3}{2}\\ \quad=-\frac{3\cos(2x+5)}{20}-\frac{\sin(2x+5)}{20}+\frac{\log3}{2}$

Hence, the general solution is;

$\displaystyle y=y_c+y_p\\ \quad=c_1e^{-x}+c_2e^{-2x}-\frac{3\cos(2x+5)}{20}-\frac{\sin(2x+5)}{20}+\frac{\log3}{2}$