First, we solve the next homogeneous differential equation $y^{\prime\prime\prime} – 2y^{\prime\prime} – y^\prime + 2y = 0$:
$\lambda^3 – 2\lambda^2 – \lambda + 2 = 0 =(\lambda-2)(\lambda-1)(\lambda+1) \Rightarrow \lambda_1 = 2, \lambda_2 = 1,\lambda_3 = -1$.
$\therefore$ The solution is $y_h(x) = c_1e^{2x}+c_2e^{x}+c_3e^{-x}$.
Now, I'm having troubles with $f(x) = \frac{2x^3+x^2-4x-6}{x^4}$, because how can I know that it has a particular form if the degree of the polynomial is negative?
Best Answer
Clearly $$f(x) = \frac{2x^3+x^2-4x-6}{x^4}=2x^{-1}+x^{-2}-4x^{-3}-6x^{-3}=(x^{-1})'''-2(x^{-1})''-(x^{-1})'+2(x^{-1})$$ which implies $y_p=x^{-1}$ is aparticular solution.