Author Topic: Q1: TUT0102  (Read 523 times)

Gavrilo Milanov Dzombeta

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Q1: TUT0102
« on: September 27, 2019, 02:01:46 PM »
$$\text{Find the general solution to the given equation by variation of parameter } $$
$$ y' + \dfrac{1}{t}y = 3\cos(2t) ,t\gt 0$$
$$\text{Make the equation homogeneous} $$
$$ y' + \dfrac{1}{t}y = 0 $$
$$\dfrac{dy}{dt} = - \dfrac{y}{t}$$
$$\dfrac{dy}{y} = - \dfrac{dt}{t} $$
$$\int \dfrac{dy}{y} = - \int \dfrac{dt}{t} $$
$$ ln(y) = - ln(t) + c $$
$$ y = e^{-ln(t)} e^{c}$$
$$ \text{let } A = e^{c} $$
$$ y = \dfrac{A}{t} \implies A(t) = yt $$
$$ y = \dfrac{A(t)}{t} $$
$$ y' + \dfrac{1}{t}y = 3\cos(2t) $$
$$ \left[\dfrac{A(t)}{t}\right]^{'} + \dfrac{1}{t^2}\left[A(t)\right] = 3\cos(2t)$$
$$ \dfrac{tA'(t) - A(t)}{t^2} +\dfrac{1}{t^2}\left[A(t)\right] = 3\cos(2t)$$
$$ \dfrac{A'(t)}{t} = 3\cos(2t) $$
$$ A'(t) = 3t\cos(2t) $$
$$ \int A'(t) dt = \int 3t\cos(2t) dt$$
$$\text{Integration by parts} $$
$$ A(t) = \dfrac{3t\sin(2t)}{2} + \dfrac{3\cos(2t)}{4} + C $$
$$ y = \dfrac{3\cos(2t)}{4t} + \dfrac{3\sin(2t)}{2t} + \dfrac{C}{t} $$