Problem 2 (afternoon)

[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]

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

[huoyanro]

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}.$

[Lan Cheng]

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)

[Changhao Jiang]

(a)
Rewrite the equation:$y''+\frac{2x+2}{x(2x+1)}y'+\frac{2}{x(2x+1)}y=0$
$p(x)=\frac{2x+2}{x(2x+1)}, W=e^{-\int p(x) dx}$
then, by Abel's theorem, $W(y_1,y_2)(x)=ce^{- \int \frac{2x+2}{x(2x+1)}}dx$
We can first calculate the integral $\int \frac{2x+2}{x(2x+1)} dx$
let $\frac{A}{x}+\frac{B}{2x+1}=\frac{2x+2}{x(2x+1)},
then \frac{(2x+1)A+Bx}{x(2x+1)}=\frac{2x+2}{x(2x+1)}$
then we get A=2, B=-2, so the integral $ \int \frac{2x+2}{x(2x+1)} dx = \int 2(\frac{1}{x}-\frac{1}{2x+1}) dx = 2(lnx-\ln(2x+1))=\ln \frac{x^2}{2x+1}$
therefore, $W=ce^{-\ln \frac{x^2}{2x+1}}=c\frac{2x+1}{x^2}$
let c=1, $W=\frac{2x+1}{x^2}$
(b)
$y_1=x+1, y_1'=1, y_1''=0$, put $y_1, y_1'$ and $y_1''$ into the equation, we can get $(2x+1)x \cdot 0+(2x+2) \cdot 1+2(x+1)=0$; therefore; $y_1$ is the solution for ODE
Let another linearly independent solution be $y_2$
Since we know $W=y_1y_2'-y_2y_1'=(x+1)y_2'-y_2$
then we need to solve the equation $(x+1)y_2'-y_2=\frac{2x+1}{x^2}$
Rewrite the equation: $y_2'-\frac{1}{x+1}y_2=\frac{2x+1}{x^2(x+1)}$
$\mu(x)=e^{\int -\frac{1}{x+1} dx}=\frac{1}{x+1}$, multiply on both sides with $\mu(x)$
We can get $\frac{1}{x+1}y_2'-\frac{1}{(x+1)^2}y_2=\frac{2x+1}{x^2(x+1)^2}$
Integrate on both sides, we can get $\frac{1}{1+x}y_2 = -\frac{1}{x(x+1)}+c$
Then $y_2 = -\frac{1}{x}+c(x+1)$, let c=1, $y_2=-\frac{1}{x}+x+1$
(c)
the general solution $y=c_1(x+1)+c_2(-\frac{1}{x}+x+1)$
since y(-1)=1, we can get $c_2=1$
$y'=c_1+c_2+\frac{c_2}{x^2}$, and y'(-1)=0, so $c_1=-2$
Therefore, $y=-(\frac{1}{x}+x+1)$

[Xuefen luo]

a)Dividing both sides by $(2x+1)x$, we have:
$y''+\frac{(2x+2)}{(2x+1)x}y'-\frac{2}{(2x+1)x}y=0$

Then, $w=ce^{-\int \frac{(2x+2)}{(2x+1)x} dx} = ce^{-\int \frac{-2}{2x+1} + \frac{2}{x} dx} = ce^{ln|2x+1|-ln|x^2|}=c((2x+1)x^{-2})$
Let $c=1, w=(2x+1)x^{-2}$.

b) $w=(x+1)y_2'-y_2=(2x+1)x^{-2}$
Then, $y_2'-\frac{1}{(x+1)}y_2=\frac{2x+1}{x^2(x+1)}$

$\mu =e^{-\int \frac{1}{x+1} dx}=e^{ln|x+1|}=\frac{1}{x+1}$

Multiplying $\frac{1}{x+1}$ to both sides of $y_2'-\frac{1}{x+1}y_2=\frac{2x+1}{x^2(x+1)}$, we have:
$\frac{1}{x+1}y_2'-\frac{1}{(x+1)^2}y_2=\frac{2x+1}{x^2(x+1)^2}$

Then, $\frac{1}{x+1}y_2=\int \frac{2x+1}{x^2(x+1)^2} dx$
$\frac{1}{x+1}y_2=\int \frac{1}{x^2}-\frac{1}{(x+1)^2} dx$
$\frac{1}{x+1}y_2=-\frac{1}{x}+\frac{1}{(x+1)^2}$
$y_2=-\frac{x+1}{x}+1$
Hence, we get $y=c_1(x+1)+c_2(-\frac{x+1}{x}+1)$

c)We have $y=c_1(x+1)+c_2(-\frac{x+1}{x}+1)$ and  $y'=c_1+\frac{1}{x^2} c_2$
Plug in $y(-1)=1,y'(-1)=0$, we get $y(-1)=c_2=1$ and $y'(-1)=c_1+c_2=0$.
Since $c_2=1,c_1=-1$.
Therefore, $y=-(x+1)-\frac{x+1}{x}+1=-x-\frac{1}{x}+1$

[Ranran Wang]

\textbf{Q2. }$(2 x+1) x y^{\prime \prime}+(2 x+2) y^{\prime}-2 y=0$.  Fnd $w\left(y_{1}, y_{2}\right)$. Check $y_{1}(x)=x+1$ is $a$ solution and fnd another linearly independent solution. Write general solution, and find solutions such fhat $y(-1)=1, \quad y^{\prime}(-1)=0$.

 \textbf{Ans:} $y^{\prime \prime}+\frac{2 x+2}{(2 x+1) x} y^{\prime}-\frac{2}{(2 x+1) x} y=0$  Fmd $P(x)=\frac{2 x+2}{(x+1) x}=\frac{A}{2 x+1}+\frac{B}{x}=\frac{A x+2 B x+B}{(2 x+1) x}$

$W=C e^{\left.-\int p( x\right) d x}=Ce^{-\int\left(\frac{-2}{2 x+1}+\frac{2}{x}\right) d x}=C e^{\int\left(\frac{2}{2 x+1}-\frac{2}{x}\right) d x}={Ce}^{\int \frac{2}{2 x+1} d x-\int \frac{2}{x} d x}=\operatorname{Ce}^{\ln| 2 x+1|-2 \ln | x |}=c \frac{e^{\ln |2 x+1|}}{e^{\sin |x|}}$

$W=\frac{2 x+1}{x^{2}}=\left|\begin{array}{ll}{y_{1}(x)} & {y_{2}(x)} \\ {y_{1}^{\prime}(x)} & {y_{2}^{\prime}(x)}\end{array}\right|=\left|\begin{array}{cc}{x+1} & {y_{2}(x)} \\ {1} & {y^{\prime}(x)}\end{array}\right|=(x+1) y_{2}^{\prime}(x)-y_{2}(x)$

$y_{1}(x)=x+1$

$y_{1}^{\prime}(x)=1$

$y_{1}^{\prime \prime}(x)=0$

$(2 x+1) x \cdot 0+(2 x+2) \cdot 1-2(x+1)=2 x+2-2 x-2=0$

$y_{2}^{\prime}(x)-\frac{1}{x+1} y_{2}(x)=\frac{2 x+1}{x^{2}(x+1)}$

$P(x)=-\frac{1}{x-4}$

$\mu=e^{\int p(x) d x}=e^{-\int \frac{1}{x+1} d x}=e^{-\ln |x+1|}=\frac{1}{x+1}$

$\frac{1}{x+1} y_{2}^{\prime}(x)-\frac{1}{(x+1)^{2}} y_{2}(x)=\frac{2 x+1}{x^{2}(x+1)^{2}}$

$\left[\frac{1}{x+1} y_{2}(x)\right]^{\prime}=\int \frac{2 x+1}{\left(x^{2}+x\right)^{2}} d x$

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

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

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

$y(-1)=1, \quad y=1, \quad x=-1$

$C_{1}(-1+1)+C_{2} \frac{-1}{-1}=1$

$C_{2}=1$

$y^{\prime}=C_{1}+\frac{C_2}{x^{-2}}$

$y^{\prime}(-1)=0$

$x=-1$

$y^{\prime}=0$

$0=C_{1}+\frac{C_2}{(-1)^{2}}=0$

$C_1+C_2=0$

$C_{1}=-1$

$y=-(x+1)+\frac{-1}{x}$

[Zuwei Zhao]

$$
(2 x+1) x y^{\prime \prime}+(2 x+2) y^{\prime}-2 y=0
$$

Find $w,$ and $y_{1}=x+1$

Find solution when $y(-1)=1 \quad y^{\prime}(-1)=0$
$$
y^{\prime \prime}+\frac{2 x+2}{(2 x+1) x} y^{\prime}-\frac{2}{(2 x+1) x} y=0
$$
$$
w=c e^{-\int \frac{2 x+2}{(2 x+1) x}} d x
$$
$$
\begin{array}{l}{\int \frac{2 x+2}{(2 x+1) x} d x} \\ {=\quad 2 \int \frac{x+1}{x(2 x+1)} \quad d x}\end{array}
$$
$$
\begin{aligned} \frac{x+1}{x(2 x+1)} &=\frac{A}{x}+\frac{B}{2 x+1} \\ x+1=& A(2 x+1)+B x \\ x+1=& A 2 x+A+B x \\ x+1=& x(2 A+B)+A \end{aligned}
$$
$$
A=1 \quad B=-1
$$
$$
\begin{aligned} \frac{x+1}{x(2 x+1)}=& \frac{1}{x}-\frac{1}{2 x+1} \\ \int \frac{x+1}{x(2 x+1)} &=\int \frac{1}{x} d x-\int \frac{1}{2 x+1} d x \\ &=\ln |x |-\int \frac{1}{2 x+1} \end{aligned}
$$
$$
\int \frac{1}{2 x+1} d x
$$
$$
\begin{aligned} & u=2 x+1 \quad \frac{d u}{d x}=2 \\=& \int \frac{1}{u} \cdot \frac{d u}{2} \\=& \frac{1}{2} \ln |u| \\=& \frac{1}{2} \ln |2 x+1| \end{aligned}
$$
$$
\begin{aligned} 2 \int \frac{x+1}{(2 x+1) x}=2 \ln |x| &-\ln |2 x+1| \\ W=\operatorname{ce}^{-\int \frac{2 x+12}{(2 x+1) x}} &=c e^{-2 \ln x+\ln 2 x+1} \\ &=c e^{-2 \ln x} \cdot e^{\ln 2 x+1} \\ &=c \frac{1}{x^{2}} \cdot 2 x+1 \end{aligned}
$$
Let $c=1$$$
=\frac{2 x+1}{x^{2}}
$$
$$
w=\left|\begin{array}{ll}{y_{1}} & {y_{2}} \\ {y_{1}^{\prime}} & {y_{2}^{\prime}}\end{array}\right|=\left|\begin{array}{cc}{x+1} & {y_{2}} \\ {1} & {y_{2}^{\prime}}\end{array}\right|=\frac{2 x+1}{x^{2}}
$$
$$
\begin{array}{l}{(x+1) y_{2}^{\prime}-y_{2}=\frac{2 x+1}{x^{2}}} \\ {y_{2}^{\prime}-\frac{1}{x+1} y_{2}=\frac{2 x+1}{x^{2}(x+1)}}\end{array}
$$
$$
\begin{array}{l}{u=e^{\int p(x) d x}=e^{-\int \frac{1}{x+1}}=e^{-\ln (x+1)}=\frac{1}{x+1}} \\ {\frac{1}{x+1} y_{2}^{\prime}-\frac{1}{(x+1)^{2}} y_{2}=\frac{2 x+1}{x^{2}(x+1)^{2}}}\end{array}
$$
$$
\begin{array}{l}{\left(\frac{1}{x+1} y_{2}\right)^{\prime}=\frac{2 x+1}{x^{2}(x+1)^{2}}} \\ {\frac{1}{x+1} y_{2}=\int \frac{2 x+1}{x^{2}(x+1)^{2}}}\end{array}
$$
$$
\begin{array}{l}{\int \frac{2 x+1}{x^{2}(x+1)^{2}} d x} \\ {\quad u=x(x+1) \frac{d u}{d x}=2 x+1} \\ {=\int \frac{1}{u^{2}} d u} \\ {=-\frac{1}{u}} \\ {=-\frac{1}{x(x+1)}+c}\end{array}
$$
$$
\begin{aligned} \frac{1}{x+1} y_{2} &=-\frac{1}{x (x+1)}+i \\ y_{2}=&-\frac{1}{x}+c(x+1) \\ &=-\frac{1}{x}+(x+1) \end{aligned} \quad \text { let } c=1
$$
$$
\begin{aligned} y=c_{1} y_{1}+c_{2} y_{2} &\left.=c_1({x}+1\right)+c_{2}\left(-\frac{1}{x}+(x+1)\right) \\ &=c_{1} x+c_{1}-\frac{c_{2}}{x}+c_{2} x+c_{2} \end{aligned}
$$
$$
\begin{array}{l}{\qquad \begin{array}{l}{y(-1)=1} \\ {1=-c_{1}+c_{1}+c_{2}-c_{2}+c_{2}} \\ {1=c_{1}} \\ {y^{\prime}=c_{1}+\frac{c_{2}}{x^{2}}+c_{2}}\end{array}}\end{array}
$$
$$
\begin{aligned} y^{\prime}(-1)&=0 \\ 0 &=c_{1}+c_{2}+c_{2} \\ &=c_{1}+2 c_{2} \end{aligned}
$$
$$
\left\{\begin{array}{l}{c_{2}=1} \\ {c_{1}+2 c_{2}=0}\end{array}\right.
$$
$$
c_{1}=-2
$$
$$
\begin{aligned} y &=-2 x-2-\frac{1}{x}+x+1 \\ &=-x-\frac{1}{x}-1 \end{aligned}
$$