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ps1sols

# ps1sols - 1 • 2 The function G of the hint satisﬁes the...

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Answers to Selected Problems in Chapter 1 1. Problem Set 1.2.1 5) This is calculus! x ( t ) = (1 / 3) t 3 - e t + 3. 6) x ( t ) = 100 e 0 . 05 t . At t = 1 this gives 105 . 13, as compared with 105 . 12 for monthly compounding. 10) If ˙ x 0 then x = γ/α . The solution for Problem 7 for arbitrary initial data x (0) = k is x ( t ) = ( k - γ/α ) e - αt + γ/α, which tends to γ/α as t → ∞ since α > 0. 12) ∂s f ( s + t ) = ∂t f ( s + t ) = f 0 ( s + t ), so the partial-differential equation reduces to 2 f 0 = αf . The solution is u ( s, t ) = C exp((1 / 2) α ( s + t )) where C is an arbitrary constant. 14) The solution is x ( t ) = x 0 exp (sin t ); it is periodic. 20) Set F ( u, v ) = R u t 0 g ( v, s ) ds . If u = u ( t ) and v = v ( t ) and f ( t ) = F ( u ( t ) , v ( t )), the chain rule gives f 0 ( t ) = ∂F ∂u u 0 ( t ) + ∂F ∂v v 0 ( t ) . Since f ( t ) = F ( t, t ) we put u ( t ) = v ( t ) = t , and obtain the stated result. 2. Problem Set 1.3.1 1) If ˙ x = k 0 x 1+ then x - (1+ ) dx dt = d dt - x - / = k 0 , so x - = C - k 0 t where C is a constant. Evaluating this gives x ( t ) = x - 0 - k 0 t - (1 / ) , which tends to infinity as t tends to t * = x - 0 / k 0 . 1

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Unformatted text preview: . 1 • 2) The function G of the hint satisﬁes the condition G ( x ,y ) = 0. Furthermore ∂G ∂y ( x ,y ) = 1 /g ( y ) 6 = 0 . Under these conditions the implicit-function theorem guarantees that there exists a unique solution y = φ ( x ) of the equation G ( x,y ) = 0 reducing to y when x = x , in a suﬃciently small neighborhood of x . • 5) The right-hand side is homogenous of degree zero and the so-lution, with y = xv , is given implicitly by Z v ds a + bs c + ds-s = ln x. • 10) Since by assumption ∂ ( pM ) ∂y = ∂ ( pN ) ∂x and ∂ ( qM ) ∂y = ∂ ( qM ) ∂y it follows immediately that the same relation holds with αp + βq in place of p or q . 2...
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