MAT244--2019F > Term Test 1

Problem 2 (afternoon)

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Victor Ivrii:
(a) Find Wronskian  $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x) , y_2(x)$ for ODE
\begin{equation*}
(2x+1)x y''+(2x+2)y'-2y=0.
\end{equation*}
(b) Check that $y_1(x)=x+1$ is a solution and find another linearly independent solution.

(c) Write the general solution, and find solution such that ${y(-1)=1, y'(-1)=0}$.

Hongling Liu:
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

Lan Cheng:
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 $y(x)=-x-1-\frac{1}{x}.$

huoyanro:
(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

huoyanro:

--- Quote from: Hongling Liu on October 23, 2019, 06:43:41 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

--- End quote ---
c should be one but not zero