Question

Using the method of undetermined coefficients, find the general solution of the differential equation y^(iv) -2y^''' +y^''=3 e^-x+2 e^x x + e^-x sin (x ) 

Answer 1

$y^{(4)}-2y^{'''}+y''=3e^{-x}+2xe^x+e^{-x}sinx$

$m^4-2m^3+m^2=0$

$m^2(m^2-2m+1)=0$

$m^2(m-1)^2=0$

$m_1=m_2=0,\ m_3=m_4=1$

The complementary solution:

$y_c=c_1+c_2x+c_3e^x+c_4xe^x$

Solve for particular solutions, using the method of undetermined coefficients:

$y_{p1}=Ae^{-x}$

$Ae^{-x}+2Ae^{-x}+Ae^{-x}=3e^{-x}$

$4A=3\implies A=3/4$

$y_{p1}=\frac{3e^{-x}}{4}$

$y_{p2}=(Ax^3+Bx^2)e^x$

$y'_{p2}=(3Ax^2+2Bx)e^x+(Ax^3+Bx^2)e^x=(Ax^3+(3A+B)x^2+2Bx)e^x$

$y''_{p2}=(3Ax^2+2(3A+B)x+2B)e^x+(Ax^3+(3A+B)x^2+2Bx)e^x=$

$=(Ax^3+(6A+B)x^2+2(3A+2B)x+2B)e^x$

$y'''_{p2}=(3Ax^2+2(6A+B)x+2(3A+2B))e^x+(Ax^3+(6A+B)x^2+$

$+2(3A+2B)x+2B)e^x=(Ax^3+(9A+B)x^2+(18A+6B)x+6A+6B)e^x$

$y^{(4)}_{p2}=(3Ax^2+2(9A+B)x+18A+6B)e^x+(Ax^3+(9A+B)x^2+$

$+(18A+6B)x+6A+6B)e^x=(Ax^3+(12A+B)x^2+(36A+8B)x+$

$+24A+12B)e^x$

$36A+8B-2(18A+6B)+6A+4B=2$

$6A=2\implies A=1/3$

$24A+12B-2(6A+6B)+2B=0$

$12A+2B=0\implies B=-2$

$y_{p2}=(\frac{x^3}{3}-2x^2)e^x$

$y_{p3}=e^{-x}(Acosx+Bsinx)$

$y'_{p3}=(Bcosx-Asinx)e^{-x}-(Acosx+Bsinx)e^{-x}=((B-A)cosx-$

$-(A+B)sinx)e^{-x}$

$y''_{p3}=((A-B)sinx-(A+B)cosx)e^{-x}-((B-A)cosx-(A+B)sinx)e^{-x}=$

$=(2Asinx-2Bcosx)e^{-x}$

$y'''_{p3}=(2Acosx+2Bsinx)e^{-x}-(2Asinx-2Bcosx)e^{-x}=$

$=((2A+2B)cosx+(2B-2A)sinx)e^{-x}$

$y^{(4)}_{p3}=((2B-2A)cosx-(2A+2B)sinx)e^{-x}-((2A+2B)cosx+$

$+(2B-2A)sinx)e^{-x}=(-4Acosx-4Bsinx)e^{-x}$

$-4A-2(2A+2B)-2B=0$

$-8A-6B=0$

$-4B-2(2B-2A)+2A=1$

$-8B+6A=1$

$A=-3B/4$

$-8B-18B/4=1\implies -50B=4\implies B=-2/25$

$A=3/50$

$y^{(4)}_{p3}=e^{-x}(\frac{3cosx}{50}-\frac{2sinx}{25})$

The general solution:

$y=y_c+y_p$

$y(x)=c_1+c_2x+c_3e^x+c_4xe^x+\frac{3e^{-x}}{4}+(\frac{x^3}{3}-2x^2)e^x+e^{-x}(\frac{3cosx}{50}-\frac{2sinx}{25})$