133_pdfsam_math 54 differential equation solutions odd

# 133_pdfsam_math 54 differential equation solutions odd -...

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Exercises 3.4 11. Let v ( t )= V [ x ( t )]. Then, using the chain rule, we get dv dt = dV dx dx dt = dV dx V and so, for V ( x ), the initial value problem (4) on page 111 of the text becomes m dV dx V = mg bV, V (0) = V [ x (0)] = v (0) = v 0 . This diFerential equation is separable. Solving yields V g ( b/m ) V dV = dx m b ± g g ( b/m ) V 1 ² dV = dx Z m b ± g g ( b/m ) V 1 ² dV = Z dx m b h mg b ln | g ( b/m ) V |− V i = x + C mg ln | mg bV | + bV = b 2 x m + C 1 . Substituting the initial condition, V (0) = v 0 , we ±nd that C 1 = mg ln | mg bv 0 | + bv 0 and hence mg ln | mg bV | + bV = b 2 x m + mg ln | mg bv 0 | + bv 0 e bV | mg bV | mg = e bv 0 | mg bv 0 | mg e b 2 x/m . 13. There are two forces acting on the shell: a constant force due to the downward pull of gravity and a force due to air resistance that acts in opposition to the motion of the shell. All of the motion occurs along a vertical axis. On this axis, we choose the origin to be the point where
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Unformatted text preview: the shell was shot from and let x ( t ) denote the position upward of the shell at time t . The forces acting on the object can be expressed in terms of this axis. The force due to gravity is F 1 = − mg, where g is the acceleration due to gravity near Earth. Note we have a minus force because our coordinate system was chosen with up as positive and gravity acts in a downward direction. The force due to air resistance is F 2 = − (0 . 1) v 2 . 129...
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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