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 October 4th, 2018, 07:07 AM #1 Senior Member   Joined: Dec 2015 From: iPhone Posts: 388 Thanks: 61 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: 7,600 Thanks: 2588 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
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 Originally Posted by v8archie 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}
This is true but only gets you one direction. He/she still has to show that if $x$ is any solution to this ODE, then it has the required form. For this it suffices to show that the homogenous equation has a unique solution (via another application of v8Archie's argument).

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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