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chapter2.page10

chapter2.page10 - = 2 y x 3 cos x(d y = 2 xy 3 x 2 e x 2(e...

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10 1. EXPLICITLY SOLVABLE FIRST ORDER DIFFERENTIAL EQUATIONS a Example 3.2 . If in Example 3.1, we set k = 0 . 2 and ω = π 4 , we would have general solution T ( t ) = e - 0 . 2 t ˆ 80 e 0 . 2 t - 5 3 . 2 0 . 64 + π 2 e 0 . 2 t cos( ωt )+ 4 π 0 . 64 + π 2 e 0 . 2 t sin( ωt ) + C ! The following graph displayed curves of several solutions with different initial values T (0) . Figure 6. Greek Letter Bar The picture above exhibits the situation when the air condition fails at t = 0, so after a out 14 hours, the room temperature is the same as the outside temperature! Practice (1) Find general solution to the following problem (a) y 0 + 4 xy = x 2 (b) e x y 0 + e x y = sin( x ) (c) xy
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Unformatted text preview: = 2 y + x 3 cos( x ) (d) y = 2 xy + 3 x 2 e x 2 (e) ( x 2 + 4) y + 3 xy = x (f) ( x 2 + 1) y + 3 x 3 y = 6 xe-3 2 x 2 (2) Application: The equation y = ky models wide range of natural phenomena–any involving a quantity whose time rate of change is proportional to its current size. Continuously Compounded Interest Rate: Let P ( t ) denote principle at time t, which will earn interest with rate r ( t ) and compounded continuously, we have dP dt = rP. Now...
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