homework 02 solutions

# homework 02 solutions - Section 1.4 25 30 d f7 = Ismxdx lny...

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Unformatted text preview: Section 1.4 25. 30. d . f7), = Ismxdx; lny = —cosx+c; y(x) = e‘c‘m‘+c = Ce'°°“ [2 = J[l+2x); lny = lnx+x2+lnC; y(x) = Cxexp(x2) y x l y(1)=1 implies C=e‘ so y(x) = xexp(x2—1). When we take square roots on both sides of the differential equation and separate variables, we get J% = I dx; J37: x—C; y(X) = (95—02- This general solution provides the parabolas illustrated in Fig. 1.4.5 in the textbook. Observe that y(x) is always nonnegative, consistent with both the square root and the original differential equation. We spot also the singular solution y(x) E O that corresponds to no value of the constant C. (a) Looking at Fig. 1.4.5, we see immediately that the differential equation ( y’)2 = 4y has no solution curve through the point (a,b) if b < 0. (b) But if b 2 0 we obviously can combine branches of parabolas with segments along the x—axis to form inﬁnitely many solution curves through (a,b) . (c) Finally, if b > 0 then on a interval containing (a,b) there are exactly two solution curves through this point, corresponding to the two indicated parabolas through (a,b) , one ascending and one descending from left to right. 31. The formal separation-of-variables process is the same as in Problem 30 where, indeed, we started by taking square roots in ( y')2 : 4 y to get the differential equation y' = 25. But whereas y' can be either positive or negative in the original equation, the latter equation requires that y’ be nonnegative. This means that only the right half of each parabola y = (x — C )2 qualiﬁes as a solution curve. Inspecting the figure , we therefore see that through the point (a,b) there passes (a) No solution curve if b < 0, (b) A unique solution curve if b > O, (c) Inﬁnitely many solution curves if b = 0, because in this case we can pick any c > a and deﬁne the solution y(x) = 0 if x _<. c, y(x) = (x— c)2 if x 2 c. —15 —10 —5 0 5 10 15 75 50 25 64. The given rate of fall of the water level is dy/dt = —4 in/hr = —(1/ 10800) ft/sec. With A = 7er and a = m2, Equation (24) is (ﬂx2)(1/10800) = —(7zr2)./2gy = —8nr2ﬁ. Hence the curve is of the form y = kx4, and in order that it pass through (1, 4) we must have k = 4. Comparing J; = 2x2 with the equation above, we see that (8r2)(10800) = 1/2, so the radius of the bottom hole is r 1/(240J3) ft = 1/35 in. Section 1.5 2 4. p=exp(_[(—2x)dx)=e'*; Dx(y-e"‘2)=1; y-e"’=x+c; y(x)=(x+C)e" 23. p = exp( 1(2 — 3/x)dx) = eZ’H'“ = x'3e2‘; Dx (y - x'3e2’) = 4e“ y-x'3e2’ =2e2" +C; y(x) = 2x3 +Cx3e'2" 29. p =exp(I(—2x)dx)=e“1; Dx (y-e""1)=e"2; y-e"‘2 =C+ fe"2 dt y(x) = e"2 (C + -§\/7?erf(x)) 38. (a) dx/dt = — x/20 and x(0) = 50 so x(t) = SOe‘mo. (b) The solution of the linear differential equation dy 5x 5y 5 4/20 1 _=_______—e ——y dt 100 200 2 40 with y(0) = 50 is y(t) = 150e-'”°—100e-”2°. (c) The maximum value of y occurs when yl(t) = _%e—1/40 +5e—t/20 = _%e—t/40 (3-4e_t/40) = 0 _ We ﬁnd that ymax= 56.251b when 1‘ 1| 401n(4/3) m 11.51 min. 40. (b) Assuming inductively that x" t"e_'/2 / (n!2”) , the equation for xn+1 iS dan 1 1 t” e'”2 __ 1 dt 2 " 2 "*‘ 1112"“ 2"” We easily solve this ﬁrst—order equation with xn+1(0) = 0 and ﬁnd that tn+1 e-I/Z x = —, (n+1)!2"+1 n+1 thereby completing the proof by induction. ...
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