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(RLC circuit) The circuit in the figure consists of a resistor $(R \text { ohms ), an inductor }(L \text { henrys), a capacitor }(C \text { farads) }$ and an inital voltage source. Let $b=R /(2 L),$ and suppose $R, L,$ and $C$ have been selected so that $b$ also equals 1$/ \sqrt{L C} .$ This is done, for instance, when the circuit is used in a voltmeter.) Let $v(t)$ be the voltage (in volts) at time $t$ measured across the capacitor. It can be shown that $v$ is in the null space $H$ of the linear transformation that maps $v(t)$into $L v^{\prime \prime}(t)+R v^{\prime}(t)+(1 / C) v(t),$ and $H$ consists of all functions of the form $v(t)=e^{-b t}\left(c_{1}+c_{2} t\right) .$ Find a basis for $H .$figure can't copy

$e^{-b t}, t e^{-b t}$ is a basis

Calculus 3

Chapter 4

Vector Spaces

Section 3

Linearly Independent Sets; Bases

Vectors

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in this problem, really given a circuit and then we're also given a the voltage to function. For Holt, it is a function of time. Since e philistines see once you two are arbitrary Dan defectors, the Judean I get to be t and t e to the negative beauty will spend the space space age Now to show that these early, nearly independent even need to show that a one some constant times experiential Bt plus eight to someone in constant times G experience you Negative. BT should be zero and a bowl ace should be zero Gosh Bebe only solution. So we're gonna try it. Try to solve this equation for different values of tea. Let's take t to be zero effort went easier. This system will be a one times one plus a two times 01 and I should be zero. So this gives us that a one is zero. Now let's take you to be one knowing that a 10 Um so we have a one, which is zero plans experiential native T plus a two times warm times exponential negative T That should be zero. This is alert Is your cause they wanted zero. Now, in order for this to be zero a two should be zero now number for the system to be equal to zero of the situation to be zero. As you can see, a one a two should be zeros and that is the only solution. So it means that these are in your independent And in the first work, we show that those actors spend space. So the basis for this system age basis for age is then exponential, negative Beatty and T experiential negativity.

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