Question

The differential equation (DE) of y'' + (5^3 + sinx)^5 y' + y = cosx^3 is?

Answer 1

$y'' + (5^3 + sinx)^5 y' + y = cosx^3$

Replace $y'$ by $D$ and y'' by $D^2$ :

$D^2+ (5^3 + sinx)^5 D+ 1 = cosx^3\\ \Rightarrow D^2+ (5^3 + sinx)^5 D+ 1- cosx^3 =0\\ \Rightarrow D=\frac{-(5^3 + sinx)^5\pm \sqrt{((5^3 + sinx)^5)^2-4(1)(1-cosx^3)}}{2}$

Here the given differential equation is of order 2 and non-homogeneous.