[정답] 공학수학 3판 zill wright
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작성일 21-10-08 02:23본문
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y − 6y + 13y = 0.
10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linear in v. However,
3. Fourth order; linear
An interval of definition for the solution of the differential equation is (−2,∞) because y is not defined at
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출판사 & 저자 : zill wright
공학수학,3판,zill
14. From y = −cos x ln(sec x + tan x) we obtain y
− 6
題目 : 공학수학 3판
= (y − x)[1 + (2(x + 2)−1/2]
-4
15. The domain of the function, found by solving x + 2 ≥ 0, is [−2,∞). From y = 1 + 2(x + 2)−1/2 we have
= tan x + cos x ln(sec x + tan x). Then y + y = tan x.
writing it in the form (v + uv − ueu)(du/dv) + u = 0, we see that it is nonlinear in u.
1.1 Definitions and Terminology
dy
7. Third order; linear
13. From y = e3x cos 2x we obtain y = 3e3x cos 2x − 2e3x sin 2x and y = 5e3x cos 2x − 12e3x sin 2x, so that
-2
− 6
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(y − x)y
5. Second order; nonlinear because of (dy/dx)2 or
dt
−20t + 20
+ 20y = 24e
6. Second order; nonlinear because of R2
x = −2.
12. From y = 6
1
[정답] 공학수학 3판 zill wright
= 24.
= y − x + 2(y − x)(x + 2)−1/2
설명
X
2. Third order; nonlinear because of (dy/dx)4
6
4
2 e−x/2. Then 2y + y = −e−x/2 + e−x/2 = 0.
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1 + (dy/dx)2
순서
레포트 > 공학,기술계열
-4 -2 2 4 t
9. Writing the differential equation in the form x(dy/dx) + y2 = 1, we see that it is nonlinear in y because of y2.
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11. From y = e−x/2 we obtain y = −1
5e
= −1 + sin x ln(sec x + tan x) and
16. Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y = 5 tan 5x is
= y − x + 8(x + 2)1/2(x + 2)−1/2 = y − x + 8.
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= y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2
y
5
1. Second order; linear
5 e−20t we obtain dy/dt = 24e−20t, so that
4. Second order; nonlinear because of cos(r + u)
5
{x
2
However, writing it in the form (y2 − 1)(dx/dy) + x = 0, we see that it is linear in x.
−20t
8. Second order; nonlinear because of x˙ 2
다.