MAT244--2019F > Term Test 1

Problem 2 (afternoon)

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

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