By Erwin Kreyszig
Read or Download Solutions manual for Advanced engineering mathematics 8ed PDF
Best mathematics books
This booklet brings jointly the most recent study achievements from sign processing and comparable disciplines, consolidating present and proposed instructions in DSP-based wisdom extraction and data fusion. The booklet comprises contributions offering either novel algorithms and latest functions, emphasizing online processing of real-world info.
- Transparent Fuzzy Systems - Modeling and Control
- Multiplier methods for mixed type equations
- Linear Algebra (3rd Edition) (Undergraduate Texts in Mathematics)
- Elementary Mathematical Models: Order Aplenty and a Glimpse of Chaos
- Elements of the Random Walk: An introduction for Advanced Students and Researchers
- Examples of the solutions of functional equations
Additional info for Solutions manual for Advanced engineering mathematics 8ed
We find ∂ ∂M = [y cos(xy)] = cos(xy) − xy sin(xy), ∂y ∂y ∂ ∂N = [x cos(xy)] = cos(xy) − xy sin(xy). ∂x ∂x Therefore, the equation is exact and F (x, y) = x cos(xy) − y −1/3 dy = sin(xy) − 3 2/3 y + h(x) 2 ∂F 2 = y cos(xy) + h (x) = √ + y cos(xy) ∂x 1 − x2 2 ⇒ h (x) = √ ⇒ h(x) = 2 arcsin x, 1 − x2 and a general solution is given by sin(xy) − 3 2/3 y + 2 arcsin x = C. 2 22. In Problem 16, we found that a general solution to this equation is exy − xy −1 = C. Substituting the initial condition, y(1) = 1, yields e − 1 = C.
Linear. 6. Linear. 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 .
2 e−x xdx = − ⇒ ⇒ ydy 1 + y2 ⇒ 2 e−x xdx = − dv u = x2 , v = 1 + y 2 v 2 = − ln |v| + C ⇒ ln 1 + y 2 − e−x = C e−u du = − −e−u is an implicit solution to the given equation. 2 18. Separating variables yields dy = 1 + y2 dy = tan xdx ⇒ 1 + y2 √ Since y(0) = 3, we have tan xdx ⇒ arctan y = − ln | cos x| + C. √ arctan 3 = − ln cos 0 + C = C ⇒ C= π . 3 Therefore, arctan y = − ln | cos x| + π 3 ⇒ y = tan − ln | cos x| + π 3 is the solution to the given initial value problem. 20. Separating variables and integrating, we get (2y + 1)dy = 3x2 + 4x + 2 dx y 2 + y = x3 + 2x2 + 2x + C.