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### Messages - huoyanro

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1
##### Term Test 1 / Re: Problem 2 (afternoon)
« on: October 23, 2019, 07:17:06 AM »

a) divide each side by $x(2x+1)$:

$y”+\frac{2x+2}{x(2x+1)}y’-\frac{2}{x(2x+1)}y=0.$

$p(x)=\frac{2x+2}{x(2x+1)},W=ce^{-\int p(x)dx}.$

let $p(x)=\frac{A}{x}+\frac{B}{2x+1}.$

$p(x)=\frac{(2A+B)x+A}{x(2x+1)}, A=2,B=-2.$

$p(x)=\frac{2}{x}-\frac{2}{2x+1}.$

$\int-p(x)dx=\int\frac{2}{2x+1}-\frac{2}{x}dx=ln(2x+1)-2ln(x).$

$W=ce^{ln(2x+1)-2ln(x)}=ce^{ln(2x+1)}e^{ln(x^{-2})}=c(2x+1)(\frac{1}{x^{2}})=c(\frac{2}{x}+\frac{1}{x^{2}})$

b) let $c=1,W=\frac{2}{x}+\frac{1}{x^{2}}=\begin{array}{cc} x+1 & y_{2}\\ 1 & y_{2}' \end{array}=(x+1)y_{2}'-y_{2}.$

divide each side by $(x+1): y_{2}'-\frac{1}{x+1}y_{2}=\frac{2}{x(x+1)}+\frac{1}{x^{2}(x+1)}.$

$\mu=e^{\int p_{2}(x)dx}=e^{\int-\frac{1}{x+1}dx}=e^{-ln(x+1)}=\frac{1}{x+1}.$

multiply each side by $\mu:\frac{1}{x+1}y_{2}'-\frac{1}{(x+1)^{2}}y_{2}=\frac{2}{x(x+1)^{2}}+\frac{1}{x^{2}(x+1)^{2}}.$

$\frac{1}{x+1}y_{2}=-\frac{1}{x(x+1)},y_{2}=-\frac{1}{x}.$

Therefore, $y_{1}=x+1,y_{2}=-\frac{1}{x}.$

c)$y(-1)=1,y'(-1)=0.$

$y(x)=C_{1}y_{1}+C_{2}y_{2}=C_{1}(x+1)+C_{2}(-\frac{1}{x}).$

$y'(x)=C_{1}+C_{2}x^{-2}.$

$\begin{cases} C_{1}=1 & C_{2}=-1\end{cases}.$

Therefore, the general solution is $y(x)=x+1+\frac{1}{x}.$
question c is wrong, C2 should be 1
because when x=-1,y=1, $y(x)=C_{1}y_{1}+C_{2}y_{2}=C_{1}(x+1)+C_{2}(-\frac{1}{x}).$
C_{2}=1\end{cases}.\$

2
##### Term Test 1 / Re: Problem 2 (afternoon)
« on: October 23, 2019, 07:10:10 AM »
2:
(2x+1)xy’’ + (2x+2)y’ -2y = 0
Solution:
a):
y’’ + [(2x+2)/(2x+1)x]y’ -[2/(2x+1)x]y= 0
W = C•e^∫-[(2x+2)/(2x+1)x]dx
W = C•(2x+1/x^2) let C = 0
W  = (2x+1/x^2)
b):
let another solution is y2
(1+x)y’2-y2 = (2x+1/x^2)
y2 = -1/x + C•(x+1)
Let C=0
y2 = -1/x + (x+1)
c):
y(x) = C1(x+1) + C2[-1/x + (x+1)]
From the problem we can get
C1 = -2
C2 = 1
∴ y(x) = -2(x+1) + [-1/x + (x+1)]
∴ y(x) = -x -(1/x) - 1
c should be one but not zero

3
##### Term Test 1 / Re: Problem 2 (afternoon)
« on: October 23, 2019, 07:06:32 AM »
(a):
y’’ + [(2x+2)/(2x+1)x]y’ -[2/(2x+1)x]y= 0
W = C•e^∫-[(2x+2)/(2x+1)x]dx
∫[(2x+2)/(2x+1)x]dx
=∫A/(2x+1) +B/x dx
B(2x+1)+Ax=2x+2
2Bx+B+Ax=2x+2
2B+A=2
B=2
A=-2
=∫-2/(2x+1) +2/x dx
=-ln(2x+1)+2ln(x)
=ln(x^(2)/(2x+1))
W = C•e^ln((2x+1/x^2)) = C•(2x+1/x^2)
let C = 1
W  = 2x+1/x^2
(b):
let another solution is y2
(1+x)y’2-y2 = 2x+1/x^2
y2 = -1/x + C•(x+1)
Let C=1
y2 = -1/x + (x+1)
c):
y(x) = C1(x+1) + C2[-1/x + (x+1)]
C1 = -2
C2 = 1
∴ y(x) = -2(x+1) + [-1/x + (x+1)]
∴ y(x) = -x -(1/x) - 1

4
##### Term Test 1 / Re: Problem 1 (afternoon)
« on: October 23, 2019, 06:57:57 AM »
(a):
My = -2y•sin(xy) - x•y^(2)•cos(xy)
Nx = -3y•sin(xy) - x•y^(2)•cos(xy)
My≠Nx
not exact
R1 = (Nx-My)/M = 1/y
u = e^(∫R1dy) = y
-y^3•sin(xy) + y•(-xy•sin(xy) + 2cos(xy) + 3y)•y’ = 0
My = -3y^2•sin(xy) - x•y^3•cos(xy)
Nx = -3y^2•sin(xy) - x•y^3•cos(xy)
My =Nx
exact
there exists ψ(x,y) such that ψx= ∫Mdx = y^2•cos(xy) + h(y)
ψy = N
ψy = 2y•cos(xy) - x•y^2sin(xy)+ h’(y)
h’(y) = 3y^2
h(y) = y^3
ψ(x,y) = y^2•cos(xy) + y^3 = C
b):
y(π/3) = 1 ∴C= 3/2
ψ(x,y) = y^2•cos(xy) + y^3 = 3/2

5
##### Term Test 1 / Re: Problem 3 (afternoon)
« on: October 23, 2019, 06:53:29 AM »
2. y'(x)= -2C1e^(-2x)-3C2e^(-3x)-2sin(2x)-10cos(2x)
C1+C2+1=0
-2C1-3C2-10=0
C1=7,C2=-8
thus y(x)=7e^(-2x)-8e^(-3x)+cos(2x)-5sin(2x)

6
##### Term Test 1 / Re: Problem 3 (afternoon)
« on: October 23, 2019, 06:48:04 AM »
1.let y'' - 5y' +6y=0
r^2-5r+6=0
r1=-1 r2=-3
thus y1(x)=C1e^(-2x) + C2e^(-3x)
let y'' - 5y' +6y= 52cos(2x)
let y2(x)=A cos(2x)+Bsin(2x)
y2'=-2Asin(2x)+2B cos(2x)
y2''=-4A cos(2x)-4Bsin(2x)
-4A cos(2x)-4Bsin(2x)+10Asin(2x)-10B cos(2x)+6A cos(2x)+6Bsin(2x)=52cos(2x)
-4A-10B+6A=52
-4B+10A+6B=0
thus A=1, B=-5
y2(x)=cos(2x)-5sin(2x)
thus y(x)=C1e^(-2x)+C2e^(-3x)+cos(2x)-5sin(2x)

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