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I've got problem with such an example:

Given equation $\frac{dx}{dt} = -x + x^7$ with initial condition $x(0) = \lambda$, where $x=x(t, \lambda)$. Find $\frac{ \partial x(t, \lambda)}{\partial \lambda} \mid _{\lambda = 0}$.

What I've got now is:

We can write: $x(t, \lambda) = \int_{0}^{t} -x + x^7 ds + C(t,\lambda)$

but since $x(0) = \lambda$, we've got $\lambda = x(0, \lambda) = \int_{0}^{0} -x + x^7 ds + C(0,\lambda) = C(0, \lambda)$ and so $x(t, \lambda) = \lambda + \int_{0}^{t} -x + x^7 ds$

Now we can apply $\frac{ \partial }{\partial \lambda}$ to both sides getting: $\frac{ \partial x(t, \lambda)}{\partial \lambda} = 1 + \int_{0}^{t}{ \frac{ \partial}{\partial \lambda}(-x(t, \lambda)) + \frac{ \partial}{\partial \lambda}(x(t, \lambda)^7) }ds$. But now I have no idea how can I calculate value of that integral.

Does anyone have idea how to move further with that solution ? Or maybe that is wrong path and there exists easier way to solve that problem?

Thanks in advance for all the help

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  • $\begingroup$ @robert-israel I get it how you've found $y(t, \lambda)$, but when I plug $\lambda=0$, nothing changes am I right ? I mean in $exp$ there will be only $0$ instead of $\lambda$. $\endgroup$
    – storaged
    Jun 30 '13 at 21:21
  • $\begingroup$ You can easily find $x(t,0)$. $\endgroup$ Jul 1 '13 at 6:55
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Let $y(t,\lambda) = \dfrac{\partial}{\partial \lambda} x(t,\lambda)$. Differentiating the differential equation $ \dfrac{\partial x}{\partial t} = - x + x^7$ with respect to $\lambda$, we get $\dfrac{\partial y}{\partial t} = - y + 7 x^6 y$, with $y(0,\lambda) = 1$. Solving this linear homogeneous initial value problem for $y$, we get $$y(t,\lambda)) = \exp\left(\int_0^s (7 x(s,\lambda)^6 - 1)\ ds\right)$$ Now plug in $\lambda=0$.

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