Math 19. Problem Set # 1 SolutionsEx. 1, 2, 4 (a, b, c, d), 5, page 43-44October 2, 2004Ex. 1a)dydt= 5yRdyy=R5dtln y = 5t + Cy = e5t+C= C · e5tb)dydt= −3yRdyy=R(−3)dtln y = −3t + Cy = e−3t+C= C · e−3tc)dydt= 12yRdyy=R12dtln y = 12t + Cy = e12t+C= C · e12td)dydt= −1.5yRdyy=R(−1.5)dtln y = −1.5t + Cy = e−1.5t+C= C · e−1.5tWhere C is a constant.You can also just use the formula for the exponential growth equation, goingdirectly fromdpdt= ap to p(t) = p(0) · eat.Thus, you can write:1a)dydt= 5yy(t) = y(0) · e5tb)dydt= −3yy(t) = y(0) · e−3tc)dydt= 12yy(t) = y(0) · e12td)dydt= −1.5yy(t) = y(0) · e−1.5tEx. 2a)y(0) = 11a) y(t) = y(0) · e5ty(t) = e5t1b) y(t) = y(0) · e−3ty(t) = e−3t1c) y(t) = y(0) · e12ty(t) = e12t1d) y(t) = y(0) · e−1.5ty(t) = e−1.5tb)y(1) = 11a) y(t) = y(0) · e5t1 = y(0) · e5·1y(0) = e−5y(t) = e−5+5t21b) y(t) = y(0) · e−3t1 = y(0) · e−3·1y(0) = e3y(t) = e3−3t1c) y(t) = y(0) · e12t1 = y(0) · e12·1y(0) = e−12y(t) = e−12+12t1d) y(t) = y(0) · e−1.5t1 = y(0) · e−1.5·1y(0) = e1.5y(t) = e1.5−1.5tc)y(−1) = 11a) y(t) = y(0) · e5t1 = y(0) · e−5·1y(0) = e5y(t) = e5+5t1b) y(t) = y(0) · e−3t1 = y(0) · e3·1y(0) = e−3y(t) = e−3−3t1c) y(t) = y(0) · e12t1 = y(0) · e−12·1y(0) = e12y(t) = e12+12t1d) y(t) = y(0) · e−1.5t1 = y(0) · e1.5·1y(0) = e−1.5y(t) = e−1.5−1.5t3d)y(−1) = −11a) y(t) = y(0) · e5t−1 = y(0) · e−5·1y(0) = − e5y(t) = −e5+5t1b) y(t) = y(0) · e−3t−1 = y(0) · e3·1y(0) = − e−3y(t) = −e−3−3t1c) y(t) = y(0) · e12t−1 = y(0) · e−12·1y(0) = − e12y(t) = −e12+12t1d) y(t) = y(0) · e−1.5t−1 = y(0) · e1.5·1y(0) = − e−1.5y(t) = −e−1.5−1.5tEx. 4 The first order Taylor’s approximation:g(x) = f (x0) + f′(x0) · (x − x0)Since x0= 0, we have: g(x) = f (0) + f′(0) · x.a)f(x) = sin(x)g(x) = sin 0 + cos 0 · x = xb)f(x) = exg(x) = e0+ e0· x = 1 + xc)f(x) =x1+x2g(x) =01+02+1+x20−x0·2x0(1+x20)2· x = 0 +11· x = x4d)f(x) = ex· sin xg(x) = e0· sin0+(ex0· cos x0+ ex0· sin x0)· x = 0 + (e0· cos 0 + e0· sin 0)x = xEx. 5Birth rate: 4/day; kbirth= 4Death rate: 1/day; kdeath= 1dPdt= (kbirth− kdeath) · PdPdt= (4 − 1)PdPdt= 3PP (t) = P (0) · e3tWe’re given in the problem that P (0) = 1000. Therefore:P (t) = 1000 ·
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