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By Florian Cajori

ISBN-10: 2102182312

ISBN-13: 9782102182310

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Y −2/3 dy = y 1/3 . 14. Since M (t, y) = et (y − t), N (t, y) = 1 + et , we find that ∂M ∂N = et = . ∂y ∂t Then F (t, y) = (1 + et )dy = (1 + et )y + h(t), ∂F = et y + h (t) = et (y − t) ∂t ⇒ h(t) = − h (t) = −tet ⇒ tet dt = −(t − 1)et , and a general solution is given by (1 + et )y − (t − 1)et = C ⇒ y= (t − 1)et + C . 1 + et 16. Computing ∂M ∂ yexy − y −1 = exy + xyexy + y −2 , = ∂y ∂y ∂N ∂ = xexy + xy −2 = exy + xyexy + y −2 , ∂x ∂x we see that the equation is exact. Therefore, F (x, y) = yexy − y −1 dx = exy − xy −1 + g(y).

8. Writing the equation in standard form, dy y − = 2x + 1, dx x we see that P (x) = − 1 x ⇒ µ(x) = exp − 1 x dx = exp (− ln x) = 1 . x 2+ 1 x dx = x (2x + ln |x| + C) . Multiplying the given equation by µ(x), we get d y 1 =2+ dx x x ⇒ y=x 41 Chapter 2 10. From the standard form of the given equation, dy 2 + y = x−4 , dx x we find that (2/x)dx = exp (2 ln x) = x2 µ(x) = exp ⇒ d x2 y = x−2 dx y = x−2 ⇒ x−2 dx = x−2 −x−1 + C = Cx − 1 . x3 12. Here, P (x) = 4, Q(x) = x2 e−4x . So, µ(x) = e4x and d 4x e y = x2 dx y = e−4x ⇒ x2 dx = e−4x x3 +C .

To separate variables, we divide the equation by x and multiply by dt. Integrating yields dx = 3t2 dt ⇒ x 3 ⇒ |x| = C2 et ln |x| = t3 + C1 ⇒ 3 +C |x| = et ⇒ 3 1 3 = eC1 et 3 x = ±C2 et = Cet , where C1 is an arbitrary constant and, therefore, C2 := eC1 is an arbitrary positive constant, C = ±C2 is any nonzero constant. Separating variables, we lost a solution 3 x ≡ 0, which can be included in the above formula by taking C = 0. Thus, x = Cet , C – arbitrary constant, is a general solution. 35 Chapter 2 12.

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A History of Mathematical Notations: Vol. I, Notations in Elementary Mathematics by Florian Cajori


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