Answer to Question #115573 in Differential Equations for Neha

Question

using the method of undetermined coefficients find the general solution of the differential equation y^iv-2y'''+2y''=3e^-x+2e^-x x+e^-x sinx

Answer 1

y’’’’ - 2*y''' + 2*y'' = 3*e^(-x) + e^(-x)*sin(x) + 2*x*e^(-x)

£^4-2£^3+2£^2=0

£^2(£^2-2£+2)=0

£=0

£^2-2£+2=0

d=4-8=-4 

£=(2+2i)/2=1+i

£=(2-2i)/2=1-i

y(1)=C1 e^(0x)+ e^(x)(C3cosx+C4sinx)=C1+ (C3cosx+C4sinx)e(x)

y(2)=Ae(-x)+Be^(-x)sin(x)+Dxe^(-x) 

y’=-Ae^(-x)+B(-e^(-x)sinx+ e^(-x)cosx)+D(-xe^(-x)+e^(-x)

y’’=Ae^(-x)+B(-2 e^(-x)cosx)+D(x-2) e^(-x)

y’’’=-A e^(-x)+2B e^(-x)(sinx+cosx) +D(3-x) e^(-x)

y’’’’=A e^(-x)-4Bsinx e^(-x)+D(x-4) e^(-x)

A e^(-x)-4Bsinx e^(-x)+D(x-4) e^(-x)-2(-A e^(-x)+2B e^(-x)(sinx+cosx) +D(3-x) e^(-x))+2(Ae^(-x)+B(-2 e^(-x)cosx)+D(x-2) e^(-x))= 3*e^(-x) + e^(-x)*sin(x) + 2*x*e^(-x)

A=688/400

B=(-25+25ctg(x))/400

D=C/e^(-x)+2/5

y(1)=C1+e^x(C3sinx+C4cosx) 

y(2)=688/400e^(-x) +1/400e^(-x)(-25sinx +25cosx)+x(C(2)+2/5e^(-x))

y=y(1)+y(2)