ODEUET2

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Consider the initial value problem $t y' - y = t 2 sin t y (t 0) = y 0$ . Show that there is a unique solution when $t_0 \ne 0$ , no solution when $t 0 = 0$ and $y_0\ne 0$ and an infinite number of solutions when $t 0 = y 0 = 0$ . Explain these reults using appropriate existence and uniqueness theorems.

$y'-\frac{1}{t}y=t\sin(t)\,$

Find the integrating factor:

$e^{-\int\frac{1}{t}dt}=e^{\ln\frac{1}{t}}=\frac{1}{t}\,$

$\frac{1}{t}y' -\frac{1}{t^2}y = \sin t\,$

$\frac{d}{dt}\left(\frac{1}{t}y\right) = \sin t\,$

$\frac{1}{t}y = -\cos t + c_1\,$

$y(t) = -t \cos t + c_1 t\,$

The inital condition gives

$y(t_0) = -t_0 \cos t_0 + c_1 t_0 = y_0\,$

$c_1 = \frac{y_0+t_0\cos t_0}{t_0} = \frac{y_0}{t_0} + \cos t_0\,$

so

$y(t) = -t\cos t + \frac{ty_0}{t_0} + t\cos t_0\,$

Main Page : Ordinary Differential Equations : PhD Qualifiers

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