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I'm currently reading about solving nonlinear differential equations, and I've stumpled upon an example that i can't seem to understand. Or rather a part of it.

Consider the system:

\begin{align} x'&=x+y^2\\ y'&=-y \end{align} Solve explicitly, the second equation yields $y(t)=y_0e^{-t}$, inserting this in the first yields \begin{align} x'=x+y^2_0e^{-2t}. \end{align} The author of the book now guesses a particular solution to be of the form $ce^{-2t}$. And inserting this guess into the equation yields the particular solution: $x(t)=-\frac{1}{3}y_0^2e^{-2t}$, he then writes that any function of the form \begin{align} x(t)=ce^t-\frac{1}{3}y_0^2e^{-2t} \end{align} And the general solution is then given by \begin{align} x(t)&=\bigg(x_0+\frac{1}{3}y_0^2\bigg)e^t-\frac{1}{3}y^2_0e^{-2t}\\ y(t)&=y_0e^{-t} \end{align}.

My question is now, where does he insert the guess $ce^{-2t}$, is it $x=ce^{-2t}$, if so, i can't seems to get the particular solution he then presents? And how does he go from the particular solution to the general solution?

A last but less important question: Is there any good litterature or method about "guessing" the particular solution?

Source: Start of Chapter 8, in Differential Equations, Dynamical Systems & An Introduction to Chaos (2nd Ed) by Morris W. Hirsch, Stephen Smale and Robert L. Devaney.

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I believe the author tacitly uses the method of undetermined coefficients. This is a method for finding particular solutions for (non-homogeneous) linear differential equations (DE) with constant coefficients.

The method works if the non-homogeneous term itself is a solution of a linear DE with constant coefficients; i.e. the non-homogeneous term is a linear combination of (co)sines, exponentials and polynomials. Note that $x'=x + y_0^2 e^{-2t}$ is such a DE.

Basically the method starts with a good educated guess (an ansatz) for the particular solution. One then fills this "candidate particular solution" in to the DE, and then solves an algebraic equation in order to find the coefficients. The link I added gives a brief description on how to make the good guess and some helpful examples.


Addendum:

As an example, lets solve the DE $x' = x + y_0^2 e^{-2t}$.

Step 1: Find the homogeneous solutions: $x_{h} = c e^t$, $c\in\mathbb{R}$.

Step 2: We want to find one particular solution $x_p$. Since the non-homogeneous term is $y_0^2 e^{-2t}$, we take a particular solution of the form $x_p = b e^{-2t}$ where $b$ is a constant we still need to determine. Plugging in this $x_p$ in the DE gives $$ \begin{align*} -2 b e^{-2t} &= b e^{-2t} + y_0^2 e^{-2t} \\ -3 b e^{-2t} &= y_0^2 e^{-2t} \\ b &= -\frac{1}{3}y_0^2. \end{align*} $$ Hence $x_p = -\frac{1}{3}y_0^2 e^{-2t}$ is a particular solution, so the general solution is $x = c e^t -\frac{1}{3}y_0^2 e^{-2t} $.

Step 3: If there are initial values, then in the last step you substitute these initial values in the general solution. It is not given explicitly in your question, but I guess that in your problem $x(0)=x_0$. Therefore $$ x_0 = x(0) = c e^0 - \frac{1}{3}y_0^2 e^{-2\cdot 0} = c - \frac{1}{3}y_0^2. $$ Therefore the solution is $$ x= \bigl(x_0 + \frac{1}{3}y_0^2\bigr) e^{t} -\frac{1}{3}y_0^2 e^{-2t}. $$

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  • $\begingroup$ Thank you, it really clarified the problem. Although i found this link (end of the comment) a bit more comprehensible, although a little less abstract i presume. cliffsnotes.com/study-guides/differential-equations/… $\endgroup$ – Jens Kramer Oct 30 '18 at 11:08
  • $\begingroup$ @Jens Kramer I indeed didn't give you any concrete example, I didn't want to make my answer bulky and unclear. But the method of undetermined coefficients is really a method one should learn by doing many examples. If you like more explanation or an example, you may ask me. $\endgroup$ – Ernie060 Oct 30 '18 at 11:11
  • $\begingroup$ Actually I'm now stuck at the fact that the general solution is $x(t)=(x_0+\frac{1}{3}y_0^2)e^t - \frac{1}{3}y_0e^{-2t}$. How does i determine $c=(x_0+\frac{1}{3}y_0^2)$? $\endgroup$ – Jens Kramer Oct 31 '18 at 10:29
  • $\begingroup$ This is not the general solution, this is a solution for the DE with initial values $x(0)=x_0$ and $y(0)=y_0$. I will add this as an example in the answer. $\endgroup$ – Ernie060 Oct 31 '18 at 10:49
  • $\begingroup$ Thanks alot for the answer, it clarified it! $\endgroup$ – Jens Kramer Oct 31 '18 at 10:57

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