공업수학 3판 (Dennis G. Zil) 솔루셥입니다.
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작성일 20-06-10 01:05본문
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공업수학 3판 (Dennis G. Zil) 솔루셥입니다. 1~5장 까지있습니다
6
4 − x2
√
5. Second order; nonlinear because of (dy/dx)2 or
레포트 > 공학,기술계열
4 − x2 = 0} or {x
(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
An interval of definition for the solution of the differential equation is (−2, 2). Other intervals are (−∞,−2)
1
공업수학 3판 (Dennis G. Zil) 솔루셥입니다.
x = π/10 + nπ/5}. From y = 25 sec2 5x we have
다운 받으시기전에 내용을 한번 읽어 보시고 내용이 맞다면 다운을 받으시기 바랍니다. 다운 받으시기전에 내용을 한번 읽어 보시고 내용이 맞다면 다운을 받으시기 바랍니다. 혹시나 받았는데 내용이 다를시 환불요청 부탁드리겠습니다.
17. The domain of the function is {x
−20t + 20
y
8. Second order; nonlinear because of x˙ 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.
An interval of definition for the solution of the differential equation is (π/2, 5π/2). Another one
X
An interval of definition for the solution of the differential equation is (−2,∞) because y is not defined at
16. Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y = 5 tan 5x is
= 24.
순서
1. Second order; linear
2
3. Fourth order; linear
= −1 + sin x ln(sec x + tan x) and
2
5x = π/2 + nπ} or {x
솔루션
-2
5
1.1 Definitions and Terminology
= (1 − sin x)−3/2 cos x = [(1 − sin x)−1/2]3 cos x = y3 cos x.
However, writing it in the form (y2 − 1)(dx/dy) + x = 0, we see that it is linear in x.
= 2xy.
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
dy
= y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2
(π/10, 3π/10), and so on.
-4
writing it in the form (v + uv − ueu)(du/dv) + u = 0, we see that it is nonlinear in u.
2. Third order; nonlinear because of (dy/dx)4
= (y − x)[1 + (2(x + 2)−1/2]
12. From y = 6
y
= tan x + cos x ln(sec x + tan x). Then y + y = tan x.
and (2,∞).
{x
10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linear in v. However,
−20t
x = π/2 + 2nπ}. From y = −1
= y − x + 2(y − x)(x + 2)−1/2
11. From y = e−x/2 we obtain y = −1
An interval of definition for the solution of the differential equation is (−π/10, π/10). Another interval is
설명
5
y − 6y + 13y = 0.
6. Second order; nonlinear because of R2
18. The function is y = 1/
x = −2.
5 e−20t we obtain dy/dt = 24e−20t, so that
+ 20y = 24e
14. From y = −cos x ln(sec x + tan x) we obtain y
− 6
− 6
= 25(1 + tan2 5x) = 25 + 25 tan2 5x = 25+y2.
7. Third order; linear
1 + (dy/dx)2
y
5e
공업수학 3판 (Dennis G. Zil) 솔루셥입니다. 혹시나 받았는데 내용이 다를시 환불요청 부탁드리겠습니다. 솔루션 보고 공부 대박나길 바랍니다.
= y − x + 8(x + 2)1/2(x + 2)−1/2 = y − x + 8.
is {x
-4 -2 2 4 t
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1 − sin x , whose domain is obtained from 1 − sin x = 0 or sin x = 1. Thus, the domain
2y
2 (1 − sin x)−3/2(−cos x) we have
dt
4
2 e−x/2. Then 2y + y = −e−x/2 + e−x/2 = 0.
= 2x
1~5장 까지있습니다
x = −2 or x = 2}. From y = 2x/(4 − x2)2 we have
1
4. Second order; nonlinear because of cos(r + u)
다. solution 보고 공부 대박나길 바랍니다.