MAT244--2019F > Term Test 1

Problem 2 (afternoon)

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huoyanro:

--- Quote from: Lan Cheng on October 23, 2019, 07:04:14 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 $y(x)=x+1+\frac{1}{x}.$

--- End quote ---
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}.$Lan Cheng: --- Quote from: Hongling Liu on October 23, 2019, 06:43:41 AM ---2: (2x+1)xy’’ + (2x+2)y’ -2y = 0 Solution: a): y’’ + [(2x+2)/(2x+1)x]y’ -[2/(2x+1)x]y= 0 W = C•e^∫-[(2x+2)/(2x+1)x]dx W = C•(2x+1/x^2) let C = 0 W = (2x+1/x^2) --- End quote --- Changhao Jiang: (a) Rewrite the equation:$y''+\frac{2x+2}{x(2x+1)}y'+\frac{2}{x(2x+1)}y=0p(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=1y'=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+1y_{1}^{\prime}(x)=1y_{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=0y_{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=-1C_{1}(-1+1)+C_{2} \frac{-1}{-1}=1C_{2}=1y^{\prime}=C_{1}+\frac{C_2}{x^{-2}}y^{\prime}(-1)=0x=-1y^{\prime}=00=C_{1}+\frac{C_2}{(-1)^{2}}=0C_1+C_2=0C_{1}=-1y=-(x+1)+\frac{-1}{x}\$