Answer to Question #151813 in Differential Equations for Komal

Question

y''-3y'+2y=3e^-x-10cosx

Answer 1

$y''-3y'+2y=3 e^x-10 \cos(x)$

$1) y''-3y'+2y=0$

$k^2-3k+2=0$

$D=9-8=1$

$k_1=\frac{3-1}{2}=1$

$k_2=\frac{3+1}{2}=2$

$Y=C_1e^x+C_2e^{2x}$

2) $y^*=(Ax+B)\cdot e^x+C\cdot cos(x)+D\cdot sin(x)$

$(y^*)'=A\cdot e^x+(Ax+B)\cdot e^x-C\cdot sin(x)+D\cdot cos(x)=$

$=(Ax+A+B)\cdot e^x-C\cdot sin(x)+D\cdot cos(x)$

$(y^*)''=(Ax+2A+B)\cdot e^x-C\cos(x)-D\cdot \sin(x). Then:$

$(Ax+2A+B)\cdot e^x-C\cos(x)-D\cdot \sin(x)-3(Ax+A+B)\cdot e^x+3C\cdot sin(x)-3D\cdot cos(x)+2(Ax+B)\cdot e^x+2C\cdot cos(x)+2D\cdot sin(x)=3e^x-10cos(x)$

$(Ax+2A+B-3Ax-3A-3B+2Ax+2B)e^x+\cos(x) (-C-3D+2C)+\sin(x)(-D+3C+2D)=3e^x-10cos(x)$

$A=-3, B=const$

$C-3D=-10, 3C+D=0 ~then ~D=3, C=-1;$

$y^*=(B-3x)e^x-\cos(x)+3\sin(x)$

$y=Y+y^*=C_1e^x+C_2e^{2x}+(B-3x)e^x-\cos(x)+3\sin(x)=$

$=e^x(C^*-3x)+C_2e^{2x}-\cos(x)+3\sin(x), C^*-const$