Question #289722

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




1
Expert's answer
2022-01-28T14:42:18-0500

The given question is;

(D2+3D+2)y=sin(2x+5)+log3\displaystyle (D^2+3D+2)y=\sin(2x+5)+\log3

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

m2+3m+2=0\displaystyle m^2+3m+2=0\\

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

Hence, the complementary function is;

yc=c1ex+c2e2x\displaystyle y_c=c_1e^{-x}+c_2e^{-2x}, where c1\displaystyle c_1 and c2\displaystyle c_2 are arbitrary constants.


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

yp=1D2+3D+2sin2(x+52)+1D2+3D+2(log3)=14+3D+2sin2(x+52)+log31D2+3D+2e0x=13D2sin2(x+52)+log32=3D+29D24sin2(x+52)+log32=3D+240sin2(x+52)+log32=140[3D(sin(2x+5))+2sin(2x+5)]+log32=140[6cos(2x+5)+2sin(2x+5)]+log32=3cos(2x+5)20sin(2x+5)20+log32\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;

y=yc+yp=c1ex+c2e2x3cos(2x+5)20sin(2x+5)20+log32\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}


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