Answer to Question #138693 in Differential Equations for Nikhil Singh

Question #138693
Find the homogeneous linear differential equation with constant coefficient that has the following function as a solution x^2e^-x +4e^x
1
Expert's answer
2020-10-15T20:00:01-0400


If the characteristic equation of linear differential equation with constant coefficient has the roots "k_1=1, k_2=k_3=k_4=-1", then its general solution is of the form:


"y=(C_1+C_2x+C_3x^2)e^{-x}+C_4e^x".


For "C_1=C_2=0, C_3=1, C_4=4" we have the particular solution "y=x^2e^{-x}+4e^x".


So, the characteristic equation is the following:


"(k-1)(k+1)^3=0" or


"(k^2\u22121)(k+1)^2=0" or


"(k^2\u22121)(k^2+2k+1)=0" or


"k^4+2k^3-2k-1=0"


Therefore, the desired differential equation is the following:


"y^{IV}+2y'''-2y'-y=0"




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