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It is said that for this type of differential equations, general solution for homogeneous case + particular solution for nonhomogeneous case = general solution for nonhomogeneous case. However, I have never seen anyone prove it and explain why. I find this to be a necessary component of the learning process. Therefore, I would appreciate it if someone could please do so. Thank you.

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  • $\begingroup$ In order to determine the one linear, I think its slope and proper translation. And Let's think solution space of ODE is a straight line. $\endgroup$ – Edgar.W Oct 24 '16 at 1:09
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$$ \sum_{i=0}^n a_i y^{(i)}(t) = f(t) \quad y = y_h + y_p $$ $$ \sum_{i=0}^n a_i (y_h+y_p)^{(i)}(t) = f(t) = \sum_{i=0}^n a_i y_h^{(i)}(t) + \sum_{i=0}^n a_i y_p^{(i)}(t) = 0+f(t)=f(t) $$

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  • $\begingroup$ $y^{(i)}$ indicates the i-th derivative (when $i=0$ means $y$) $\endgroup$ – Anonymous Oct 24 '16 at 2:33
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Consider the general $n$-th order constant-coefficient inhomogeneous linear ODE, $$y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_1y' + a_0y = f.$$ Suppose $y_1$ and $y_2$ are two particular solutions to this equation. Then $y_1 - y_2$ solves the homogeneous equation $$y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_1y' + a_0y = 0.$$ The general solution to this equation depends on $n$ arbitrary constants $C_1,\ldots,C_n$. Let $y_h(t;C_1,\ldots,C_n)$ denote the solution corresponding to these constants. Then since $y_1-y_2$ solves the homogeneous equation, there exists a choice of coefficients $C_1,\ldots,C_n$ such that $y_1(t) - y_2(t) = y_h(t;C_1,\ldots,C_n)$. Then it follows that $$y_1(t) = y_2(t) + y_h(t;C_1,\ldots,C_n).$$ Thus $y_1$ is a sum of a particular solution of the inhomogeneous equation and a solution of the homogeneous equation; moreover, since $y_1$ and $y_2$ were arbitrary solutions of the inhomogeneous equation, we have shown that every pair of solutions has this property.

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