Question : Find a homogeneous linear differential equation with constant coefficients which is satisfied by the function y = 1 + x + cos^3 x
Solution :
Given function
\[y\ =\ 1\ +\ x\ +\ cos^3\ \ x \] (1)
Differentiating this function with it with respect to x
\[\frac{dy}{dx}=-3sin\left(x\right)\cos^2{\left(x\right)}+1 \] (2)
\[\frac{d^2y}{dx^2}=6\sin^2{(x)}\cos(x)-3\cos^3{(x)} \] (3)
\[ \frac{d^3y}{dx^3}=-6\sin^3{\left(x\right)}+21sin\left(x\right)\cos^2{(x)} \] (4)
\[\frac{d^4y}{dx^4}=-60\sin^2{\left(x\right)}cos(x)+21\cos^3{(x)} \] (5)
\[\frac{d^5y}{dx^5}=60\sin^3{(x)}-183sin\left(x\right)\cos^2{(x)} \] (6)
\[\frac{d^6y}{dx^6}=546\sin^2{\left(x\right)}cos\left(x\right)-183\cos^3{\left(x\right)} \] (7)
From equations (3), (5)
\[10\frac{d^2y}{dx^2}+\frac{d^4y}{dx^4}=-9cos^3x , \frac{d^4y}{dx^4}+7\frac{d^2y}{dx^2}=-18\sin^2{(x)}cos(x)\]Therefore
\[18\ast\frac{d^6y}{dx^6}+180\ast\frac{d^4y}{dx^4}+162\ast\frac{d^2y}{dx^2}+0\frac{dy}{dx}+0y=0\]Is the required homogeneous linear differential equation with constant coefficients which is satisfied by the function y = 1 + x + cos^3 x