Question #149011
Find a homogeneous linear differential equation with constant coefficient that is satisfied by y=4xe^xsin2x
1
Expert's answer
2020-12-07T20:08:07-0500

y=4xexsin(2x)Ify=eαx(Acos(βx)+Bsin(βx)),the auxiliary equation ism=α±jβm=1±j2The quadratic equation whose roots are1±j2arem2(1+j2+1j2)m+(1+j2)(1j2)=0m22m+5=0Forexsin(2x)to be multiplied byx,the roots ofmmust be repeated.m=1±j2twice(m22m+5)2=0m44m3+14m220m+25=0The homogenous differential equation isy(iv)4y(iii)+14y20y+25y=0\displaystyle y = 4xe^x\sin(2x)\\ \textsf{If}\,\, y = e^{\alpha x}(A\cos(\beta x) + B\sin(\beta x)),\\ \textsf{the auxiliary equation is}\\ m = \alpha \pm j\beta\\ \therefore m = 1 \pm j2\\ \textsf{The quadratic equation whose roots are}\,\, 1 \pm j2 \,\, \textsf{are}\\ m^2 - (1 + j2 + 1 - j2)m + (1 + j2)(1 - j2) = 0\\ m^2 - 2m + 5 = 0\\ \textsf{For}\,\, e^x\sin(2x) \,\, \textsf{to be multiplied by}\,\, x, \\ \textsf{the roots of}\,\,m\,\,\textsf{must be repeated.}\\ \therefore m = 1 \pm j2 \,\,\textsf{twice}\\ (m^2 - 2m + 5)^2 = 0\\ m^4 - 4m^3 + 14m^2 - 20m + 25 = 0\\ \therefore \textsf{The homogenous differential equation is}\\ y^{(iv)} - 4y^{(iii)} + 14y'' - 20y' + 25y = 0


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