EE102 Systems and Signals (Discussion)

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General Case

The general first-order linear differential equation is given by:

$$\frac{dy(t)}{dt} + a(t)y(t) = x(t), y(0) = y_o$$

where a(t) and x(t) are given functions, and we need to solve for $y(t)$ for $t\geq 0$ given the initial condition $y(0) = y_o$.

To solve the equation for $y(t)$, we use the well-known trick of multiplying the equation by the so-called integrating factor:

$$e^{\int_0^t a(\tau) d\tau}$$

This gives:

$$e^{\int_0^t a(\tau) d\tau}\frac{dy(t)}{dt} + e^{\int_0^t a(\tau) d\tau}a(t)y(t) = e^{\int_0^t a(\tau) d\tau}x(t)$$

But by the product rule of differentiation, the LHS is

$$\frac{d}{dt}\left( e^{\int_0^t a(\tau)d\tau}y(t)\right)$$

since

$$\frac{d}{dt}\left( e^{\int_0^t a(\tau)d\tau}y(t)\right) = e^{\int... ...frac{dy(t)}{dt} + \frac{d}{dt}\left( e^{\int_0^t a(\tau)d\tau}\right)\cdot y(t)$$     (product rule)$$$$

and

$$\frac{d}{dt}\left(e^{\int_0^t a(\tau) d\tau}\right) = e^{\int_0^t... ...rac{d}{dt}\left(\int_0^t a(\tau) d\tau\right) = e^{\int_0^t a(\tau) d\tau}a(t)$$

Hence, the differential equation now becomes:

$$\frac{d}{dt}\left( e^{\int_0^t a(\tau)d\tau}y(t)\right) = e^{\int_0^t a(\tau) d\tau}x(t)$$

To get $y(t)$, we first change the $t$ to $\sigma$ and integrate from 0 to $t$:

$$\int_0^t\frac{d}{d\sigma}\left( e^{\int_0^\sigma a(\tau)d\tau}y(\sigma)\right) d\sigma = \int_0^te^{\int_0^\sigma a(\tau) d\tau}x(\sigma) d\sigma$$

The integral on the LHS is just $e^{\int_0^\sigma a(\tau)d\tau}y(\sigma)$. Therefore, using the initial condition $y(0) = y_o$, we get

$$\left[ e^{\int_0^\sigma a(\tau)d\tau}y(\sigma)\right]_{\sigma=0}^... ...(\tau) d\tau} - y_o = \int_0^te^{\int_0^\sigma a(\tau) d\tau}x(\sigma) d\sigma$$

Finally, rearranging terms, we get the solution to be

$$y(t) = y_oe^{-\int_0^t a(\tau)d\tau} + e^{-\int_0^t a(\tau)d\tau} \int_0^te^{\int_0^\sigma a(\tau) d\tau}x(\sigma) d\sigma,   t \geq 0$$

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