공업수학 3판 (Dennis G. Zil) 솔루셥입니다.
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작성일 20-05-26 19:57본문
Download : 최신공업수학.zip
However, writing it in the form (y2 − 1)(dx/dy) + x = 0, we see that it is linear in x.
4 − x2 = 0} or {x
설명
1 + (dy/dx)2
2
17. The domain of the function is {x
√
y − 6y + 13y = 0.
다운 받으시기전에 내용을 한번 읽어 보시고 내용이 맞다면 다운을 받으시기 바랍니다. = 24.
2. Third order; nonlinear because of (dy/dx)4
5. Second order; nonlinear because of (dy/dx)2 or
1~5장 까지있습니다
x = −2.
= y − x + 8(x + 2)1/2(x + 2)−1/2 = y − x + 8.
공업수학 3판 (Dennis G. Zil) 솔루셥입니다.
레포트 > 공학,기술계열
10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linear in v. However,
= (1 − sin x)−3/2 cos x = [(1 − sin x)−1/2]3 cos x = y3 cos x.
y
18. The function is y = 1/
12. From y = 6
1 − sin x , whose domain is obtained from 1 − sin x = 0 or sin x = 1. Thus, the domain
x = π/2 + 2nπ}. From y = −1
5
11. From y = e−x/2 we obtain y = −1
5e
1
−20t
2y
− 6
8. Second order; nonlinear because of x˙ 2
+ 20y = 24e
6. Second order; nonlinear because of R2
− 6
An interval of definition for the solution of the differential equation is (−2,∞) because y is not defined at
4. Second order; nonlinear because of cos(r + u)
and (2,∞).
1. Second order; linear
-4
-2
16. Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y = 5 tan 5x is
{x
공업수학 3판 (Dennis G. Zil) 솔루셥입니다. 혹시나 받았는데 내용이 다를시 환불요청 부탁드리겠습니다. 해결책 보고 공부 대박나길 바랍니다. 혹시나 받았는데 내용이 다를시 환불요청 부탁드리겠습니다.
4 − x2
2 e−x/2. Then 2y + y = −e−x/2 + e−x/2 = 0.
= y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2
9. Writing the differential equation in the form x(dy/dx) + y2 = 1, we see that it is nonlinear in y because of y2.
= y − x + 2(y − x)(x + 2)−1/2
= −1 + sin x ln(sec x + tan x) and
5
x = −2 or x = 2}. From y = 2x/(4 − x2)2 we have
writing it in the form (v + uv − ueu)(du/dv) + u = 0, we see that it is nonlinear in u.
= 2x
= tan x + cos x ln(sec x + tan x). Then y + y = tan x.
(π/10, 3π/10), and so on.
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순서
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
y
7. Third order; linear
4
An interval of definition for the solution of the differential equation is (π/2, 5π/2). Another one
5 e−20t we obtain dy/dt = 24e−20t, so that
An interval of definition for the solution of the differential equation is (−π/10, π/10). Another interval is
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y
1.1 Definitions and Terminology
dy
= (y − x)[1 + (2(x + 2)−1/2]
−20t + 20
x = π/10 + nπ/5}. From y = 25 sec2 5x we have
(y − x)y
15. The domain of the function, found by solving x + 2 ≥ 0, is [−2,∞). From y = 1 + 2(x + 2)−1/2 we have
= 25(1 + tan2 5x) = 25 + 25 tan2 5x = 25+y2.
An interval of definition for the solution of the differential equation is (−2, 2). Other intervals are (−∞,−2)
-4 -2 2 4 t
6
2 (1 − sin x)−3/2(−cos x) we have
1
dt
다. 다운 받으시기전에 내용을 한번 읽어 보시고 내용이 맞다면 다운을 받으시기 바랍니다. 1~5장 까지있습니다
3. Fourth order; linear
5x = π/2 + nπ} or {x
2
14. From y = −cos x ln(sec x + tan x) we obtain y
is {x
= 2xy.
X
공업수학 3판 (Dennis G. Zil) 솔루셥입니다. 솔루션 보고 공부 대박나길 바랍니다.