Answer to Question #145274 in Differential Equations for Abhijeet Pundir

Question

Y^4-y=x

Answer 1

First we solve $y^{(4)}-y=0$ .

$k^4-1=0, k_1=1, k_2=-1, k_3=i, k_4=-i$ . These numbers correspond to function $e^x, e^{-x},cos(x), sin(x)$ . So the solution is $y=c_1e^x+c_2e^{-x}+c_3cos(x)+c_4sin(x)$ .

Since the right part of the non-zero equation is x then we'll need a first power polynomial to add to zero equation's solution (Ax+B):

$(Ax+B)^{(4)}-(Ax+B)=x$

$-A=1, A=-1$

So the solution of the non-zero equation is

$y=c_1e^x+c_2e^{-x}+c_3cos(x)+c_4sin(x)-x$