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October 4th, 2018, 07:07 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 271 Thanks: 31  Homogeneous solution explain
Given equation $\displaystyle y'+py=q$ for $\displaystyle q=0$ then $\displaystyle y=y_h$ is the homogeneous solution (1) Explain why solution to equation is $\displaystyle y=y_h+y_p$ where $\displaystyle y_h$  homogeneous and $\displaystyle y_p$ particular (2) Explain how to derive $\displaystyle y_p $ from $\displaystyle y_h$ How to derive $\displaystyle y=y_h + y_p$ ? 
October 4th, 2018, 12:11 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,511 Thanks: 2514 Math Focus: Mainly analysis and algebra 
1) Suppose that $y_p$ is any solution of $y'+py=q$ and that $y_c$ is any solution of $y'+py=0$. Then if $y= y_c+y_p$, we have $y'=y_c'+y_p'$ and so \begin{align}y'+py &= (y_c'+y_p') + p(y_c+y_p) \\ &= (y_c' + py_c) + (y_p' + py_p) &(\text{just grouping the terms differently}) \\ &= 0 + q \\ &= q\end{align}

October 4th, 2018, 01:03 PM  #3  
Senior Member Joined: Sep 2016 From: USA Posts: 520 Thanks: 293 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
For this the usual approach is to assume write the equation as $x' = Ax$ so that $\exp(tA)$ is a solution. Now assume that $x$ is a solution and and show that $\exp(tA)x(t)$ is constant. For the second part, this is just the variation of constants formula: \[x(t) = \exp(tA) \left( \exp(t_0A)x_0 + \int_{t_0}^{t} \exp(sA) g(s) \ ds \right) \]  

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