Dependence on Parameters
Consider the differential equation
\begin{equation}
\dot x = f(t, x, \alpha),\qquad x(0) = x_0,
\label{eq-ivp-with-parameter}
\end{equation}
where $\alpha$ is a parameter (i.e. a constant that appears
in the differential equation).
We assume that the function $f$ is a continuously differentiable function of
$(t, x, \alpha)$.
Theorem. Assume that for some choice $(\bar \alpha, \bar x_0)$ of the
parameter and initial value, the initial value problem
\eqref{eq-ivp-with-parameter} has a solution $\bar x(t)$ that is defined on a
closed interval $[0, t_1]$. Then there is an $\varepsilon\gt0$ such that for
all $\alpha$ and $x_0$ with $|\alpha-\bar \alpha| \lt \varepsilon$ and
$|x_0-\bar x_0|\lt \varepsilon$ the equation \eqref{eq-ivp-with-parameter}
has a solution $x(t, \alpha, x_0)$ that is defined on the same interval $[0,
t_1]$. The equation $x(t, \alpha, x_0)$ is a continuously differentiable
function of $(t, \alpha, x_0)$.
The partial derivatives with respect to the parameters $\alpha$ and
$x_0$ satisfy their own differential equation. This equation is
easily derived by “differentiating the equation”. For example, $y=\frac{\pd x} {\pd \alpha}$ satisfies
\[
\frac{\pd y} {\pd t} = f_x(t, x(t), \alpha) y(t) + f_\alpha(t, x(t), \alpha), \qquad
y(0) = 0,
\]
while $z = \frac{\pd x} {\pd x_0}$ satisfies
\[
\frac{\pd z} {\pd t} = f_x(t, x(t), \alpha) z(t), \qquad z(0) = 1.
\]
Each of these two equations is called “the variational equation” for the differential equation \eqref{eq-ivp-with-parameter}.
The variational equation is always a first order linear differential
equation: the equation for $y$ is inhomogeneous, the equation for $z(t)$ is
homogeneous.
For a review of how to solve linear first order equations, follow this link.
Example: solving the variational equation
Let $x(t, \alpha)$ be the solution of
\[
\dot x = x(1-x) + \alpha\sin kt, \qquad x(0)=0.
\]
Here $k$ and $\alpha$ both are constants, but we will only vary $\alpha$ (in
this example—you could also consider variations in $k$.)
For general $\alpha$ we do not have a formula for the solution to this
equation, but for $\alpha=0$ we do have one, namely: $x(t, 0) = 0$.
For small values of $\alpha$ the solution should be approximately given by
\begin{equation}
x(t, \alpha) = x(t, 0) + \frac{\pd x}{\pd \alpha}(t, 0)\,\alpha +\dots
\label{eq-Taylor-in-alpha}
\end{equation}
by Taylor’s formula (the dots “$\dots$” contain the higher
order terms, and the remainder term). Here we have used Taylor’s formula
in the following way. Freeze the variable $t$ at some value, and think of
$x(t, \alpha)$ as a function of $\alpha$ only. Taylor then says
\[
x(t, \alpha) = x(t, 0) + \frac{\pd x}{\pd\alpha}(t,0) \alpha
+\frac{\pd^2 x}{\pd\alpha^2}(t,0) \frac{\alpha^2}{2!}
+\frac{\pd^3 x}{\pd\alpha^3}(t,0) \frac{\alpha^3}{3!}
+\cdots
\]
The derivative
\[
y(t) = \frac{\pd x}{\pd \alpha}(t, 0)
\]
satisfies the equation
\[
\dot y = \bigl(1-2x(t, 0)\bigr) y(t) + \sin kt, \qquad y(0)=0
\]
or, taking into account that our known solution is $x(t, 0) = 0$,
\[
\dot y = y + \sin kt, \quad y(0)=0.
\]
The solution of this linear equation is (see
below for the details)
\[
y(t)
= \frac{-\sin kt- k\cos kt + k e^{t}}{1+k^2}.
\]
Conclusion: for small values of $\alpha$ the solution $x(t, \alpha)$ is
approximately given by the Taylor expansion \eqref{eq-Taylor-in-alpha}
\[
x(t, \alpha)
= \alpha y(t) + \cdots
= \alpha \frac{k e^{t} -\sin kt- k\cos kt}{1+k^2} + \cdots
\]
Solving that linear equation
Rewrite it as
\[
\dot y - y = \sin kt
\]
and multiply both sides with the integrating factor $e^{-t}$:
\[
\frac{d}{dt}\bigl(e^{-t}y\bigr)
= e^{-t}\dot y - e^{-t}y
= e^{-t}\sin kt.
\]
Now integrate $\frac{d}{dt}(e^{-t}y)$ to get $e^{-t}y$:
\[
e^{-t}y = \int e^{-t}\sin kt \, dt
= - e^{-t} \frac{\sin kt + k\cos kt}{1+k^2} + C.
\]
Don’t forget the “$+C$”'! (How do you do the integral?
Integrate by parts twice, look it up, or ask Wolfram).
The initial condition $y(0)=0$ tells us that
\[
0 = -\frac{k}{1+k^2} + C \implies C=\frac{k}{1+k^2}.
\]
Therefore the solution is
\[
y(t) = - \frac{\sin kt+ k\cos kt}{1+k^2} + C e^t
= \frac{-\sin kt- k\cos kt + k e^{t}}{1+k^2}.
\]