Ian Chen 發問時間： 科學數學 · 1 0 年前

# -math prove-->

suppose a function y=f(x) satisfies the differential equation dy=4ysin2xdx and the initial condition y(π)=e the purpose of this exercise is to find y(π/6).

A. separate variables and integrate to obtain lny= c-2cos2x, y>0. by use of the initial condition show that c=3, and then get y(π/6).

B. the initial condition y(π)-e means that x-πcorresponds to y-e. likewise, x-(π/6)corresponds to y=a, where a=y(π/6). integrating between corresponding limits gives ∫上a 下e dy/y=∫上π/6 下π 4sin2xdx. evaluate the definite integrals and solve the resulting equation for a.

C. if xdy +3ydx=0 and y(-π)=e you cannot find y(π). why not?

### 1 個解答

• 天助
Lv 7
1 0 年前
最佳解答

QA:(indefinite integration)

(1/y)dy/dx=4sin(2x) (y>0)

∫ (1/y) (dy/dx) dx = ∫ 4sin(2x) dx

(by changing variable or substitution) ∫(1/y) dy= ∫4sin(2x) dx

Ln(y)= c-2cos(2x)

y(π)=e (x=π, y=e), then 1=c-2, c=3, hence, Ln(y)=3-2cos(2x) (for y>0)

x=π/6, Ln(y)=2, thus y=e^2

QB:(definite integration)

∫[π~π/6] (1/y) (dy/dx) dx = ∫[π~π/6] 4sin(2x) dx

∫[e~a] (1/y)dy= ∫[π~π/6] 4sin(2x) dx

Ln(a)-1=-2cos(2x) sub. for x=π~π/6

Ln(a)=1-1+2=2, a=y(π/6)=e^2

QC:

xdy+3ydx=0, y(-π)=e, let y(π)=a

(indefinite integration)

∫ (1/y)(dy/dx) dx=∫ 1/y dy= -∫ 3/x dx

Ln|y|= -3Ln|x|+c ----(A)

x=-π, y=e, 1=-3Ln(π)+c, c=1+3Ln(π)

x=π, y=?, Ln|y|= -3Ln(π)+c (for unknown constant c)

(Note: the constant c in the expression (A) can be determined in the

neighbourhood of the point (-π, e) only, not the whole plane)

(definite integration)

∫[e~a] (1/y)(dy/dx) dx= ∫[-π~π] -3/x dx

the right integral is not defined, so we cannot obtain the value a=y(π).

Note: In the problem QA,QB, dy/dx= 4sin(2x) is diff. for any x;

while QC, dy/dx= -3y/x is not diff. for any x. (cannot cross the origin.)