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An IVP of the form

\begin{align} x'=p(t)x+q(t) \hspace{.25em}; \hspace{.5em}x(t_0)=x_0 \end{align}

has a general solution as follows

\begin{align} \displaystyle x(t)=e^{\int_{t_0}^{t}p(\tau)d\tau}x_0+e^{\int_{t_0}^{t}p(\tau)d\tau}\int_{t_0}^te^{-\int_{t_0}^{s}p(\tau)d\tau}q(s)ds \end{align} I can easily reach this conclusion but only with indefinite integrals because the integration constant yields the first term after the equal sign; however, the above solution was presented to me and I am not sure where the definite integrals came from, any thoughts?

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Let $f:A\rightarrow B$ be an integrable function, with the anti derivative being $F.$ Let $x_0\in A$ and $c\in B$.

You can write an indefinite integral say $$\int f(x) dx=F(x)+c$$ as something definite like $$\int_{x_0}^{x}f(\alpha)d\alpha=F(x)-F(x_0).$$ Now you could choose $c=-F(x_0)$. So they are essentially the same thing.

We can solve your problem using the integration factor $$e^{-\int_{t_0}^{t}p(\alpha) d\alpha}.$$

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Since $p(x)$ and $q(x)$ are unknown the indefinite integrals can't be evaluated to some close form. So general solution has to stay in integral form. The constant of integration is evaluated by initial condition and "re-combine" with the indefinite integral to form a definite integral. Definite integral is there to satisfy initial condition.

The general solution of the first order differential equation without applying its initial condition can be written as $$x(t)=c \cdot e^{A(t)} + e^{A(t)} \cdot B(t)$$ where $A(t)=\int p(t)dt$ and $B(t)=\int q(t) \cdot e^{-A(t)}dt$. Lastly, $c$ is the integration constant to be determined by initial condition of differential equation. Applying initial condition we have: $c={x_0} \cdot e^{-A(t_0)} - B(t_0) $. Thus the general solution becomes $$x(t)=x_0 \cdot e^{A(t)-A(t_0)} + e^{A(t)} \cdot (B(t)-B(t_0)) .$$ Some of solution's terms can be written in definite integral form: $$x(t)=x_0 \cdot e^{\int _{t_0}^{t}p(\tau)d\tau} + e^{\int _{t_0}^{t}p(\tau)d\tau} \cdot e^{A(t_0)} \cdot \int _{t_0}^{t}q(\tau) e^{-\int p(\tau)d\tau} d\tau .$$ By moving constant $e^{A(t_0)}$ in the third integral we have $$x(t)=x_0 \cdot e^{\int _{t_0}^{t}p(\tau)d\tau} + e^{\int _{t_0}^{t}p(\tau)d\tau} \cdot \int _{t_0}^{t}q(\tau) e^{A(t_0)-\int p(\tau)d\tau} d\tau .$$ Applying definition of definite integral to fourth integral, we get $$x(t)=x_0 \cdot e^{\int _{t_0}^{t}p(\tau)d\tau} + e^{\int _{t_0}^{t}p(\tau)d\tau} \cdot \int _{t_0}^{t}q(\tau) e^{-\int _{t_0}^{\tau} p(s)ds} d\tau .$$ Q.E.D.

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