![]() ![]() vector calculus matthews pdf MathStudy/Calculus.md at master We may rewrite Equation (1.13) using indices as. As the set fe^ igforms a basis for R3, the vector A may be written as a linear combination of the e^ i: A= A 1e^ 1 + A 2e^ 2 + A 3e^ 3: (1.13) The three numbers A i, i= 1 2 3, are called the (Cartesian) components of the vector A. 1.2 Vector Components and Dummy Indices Let Abe a vector in R3. The domain of the function, found by solving x + 2 ≥ 0, is = y − x + 2(y − x)(x + 2) −1/2 = y − x + 2(x + 2) −1/2 = y − x + 8(x + 2) 1/2 (x + 2) −1/2 = y − x + 8.Vector calculus textbook solutions pdfLectures on Vector Calculus - CSUSB. From y = − cos x ln(sec x + tan x) we obtain y = −1 + sin x ln(sec x + tan x) and y = tan x + cos x ln(sec x + tan x). From y = e 3x cos 2x we obtain y = 3e 3x cos 2x − 2e 3x sin 2x and y = 5e 3x cos 2x − 12e 3x sin 2x, so that y − 6y + 13y = 0. From y = e −x/2 we obtain y = − 1 2 e −x/2. ![]() However, writing it in the form (v + uv − ue u)(du/dv) + u = 0, we see that it is nonlinear in u. Writing the differential equation in the form u(dv/du) + (1 + u)v = ue u we see that it is linear in v. However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is linear in x. Writing the differential equation in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in y because of y 2. Second order nonlinear because of ˙ x 2 9. Second order nonlinear because of (dy/dx) 2 or 1 + (dy/dx) 2 6. Second order nonlinear because of cos(r + u) 5. Third order nonlinear because of (dy/dx) 4 3. The graph of the solution defined on (−∞, ln 2) is dashed, and the graph of the solution defined on (ln 2, ∞) is solid.ġ. Thus, the solution is defined on (−∞, ln 2) or on (ln 2, ∞). Exponentiating both sides of the implicit solution we obtain 2X − 1 X − 1 = e t =⇒ 2X − 1 = Xe t − e t =⇒ (e t − 1) = (e t − 2)X =⇒ X = e t − 1 e t − 2. ![]() Second-order nonlinear because of ˙ x 2 11. Second-order nonlinear because of 1/R 2 9. Second-order nonlinear because of 1 + (dy/dx) 2 8. Second-order nonlinear because of cos(r + u) 7. Writing it in the form u(dv/du) + (1 + u)v = ue u we see that it is linear in v. The differential equation is first-order. Writing it in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in y because of y 2. Third-order nonlinear because of (dy/dx) 4. ![]()
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