Answer to Question #138693 in Differential Equations for Nikhil Singh

Question

Find the homogeneous linear differential equation with constant coefficient that has the following function as a solution x^2e^-x +4e^x

Answer 1

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

$(k^2−1)(k+1)^2=0$ or

$(k^2−1)(k^2+2k+1)=0$ or

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

Therefore, the desired differential equation is the following:

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