[해법] 工學(공학) 수학 3판 zill wright
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작성일 19-10-07 12:23
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X
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 − x + 8(x + 2)1/2(x + 2)−1/2 = y − x + 8.
5. Second order; nonlinear because of (dy/dx)2 or
5
레포트 > 공학,기술계열
−20t
공학수학,3판,zill
1. Second order; linear
タイトル : 工學(공학) 수학 3판
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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
5 e−20t we obtain dy/dt = 24e−20t, so that
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2. Third order; nonlinear because of (dy/dx)4
= 24.
(y − x)y
= y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2
14. From y = −cos x ln(sec x + tan x) we obtain y
4
7. Third order; linear
-2
10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linear in v. However,
= (y − x)[1 + (2(x + 2)−1/2]
dt
1
순서
12. From y = 6
출판사 & 저자 : zill wright
= tan x + cos x ln(sec x + tan x). Then y + y = tan x.
참고 : 1~20챕터까지
-4 -2 2 4 t
dy
9. Writing the differential equation in the form x(dy/dx) + y2 = 1, we see that it is nonlinear in y because of y2.
x = −2.
6
+ 20y = 24e
3. Fourth order; linear
5e
2
16. Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y = 5 tan 5x is
y − 6y + 13y = 0.
설명
However, writing it in the form (y2 − 1)(dx/dy) + x = 0, we see that it is linear in x.
2 e−x/2. Then 2y + y = −e−x/2 + e−x/2 = 0.
−20t + 20
6. Second order; nonlinear because of R2
Download : 공업수학[dennisg[1].zip( 64 )
8. Second order; nonlinear because of x˙ 2
4. Second order; nonlinear because of cos(r + u)
-4
[해법] 工學(공학) 수학 3판 zill wright
제목 : 공학수학 3판 출판사 & 저자 : zill wright 참고 : 1~20챕터까지 잘못된 자료가 있을시 환불요청 하시면 환불을 받으실수 있습니다. An interval of definition for the solution of the differential equation is (−2,∞) because y is not defined at
{x
1 + (dy/dx)2
= y − x + 2(y − x)(x + 2)−1/2
− 6
5
1.1 Definitions and Terminology
writing it in the form (v + uv − ueu)(du/dv) + u = 0, we see that it is nonlinear in u.
= −1 + sin x ln(sec x + tan x) and
11. From y = e−x/2 we obtain y = −1
− 6
잘못된 data(資料)가 있을시 환불요청 하시면 환불을 받으실수 있습니다.