Toronto Math Forum

MAT244--2019F => MAT244--Test & Quizzes => Quiz-1 => Topic started by: suyichen on September 27, 2019, 02:00:01 PM

Title: TUT0801
Post by: suyichen on September 27, 2019, 02:00:01 PM
Show that it is the homogenous equation and find
the general solution of:
$$
\begin{array}{l}{\quad \frac{d y}{d x}=\frac{x+3 y}{x-y}} \\ {\frac{d y}{d x}=\frac{x+3 y}{x-y}=\frac{1+3\left(\frac{x}{x}\right)}{1-\frac{y}{x}}}\end{array}
$$
$$
\begin{array}{l}{\text { Let } v=\frac{y}{x} \quad y=v x} \\ {\frac{d y}{d x}=\frac{d v}{d x} x+v \cdot 1} \\ {\frac{d v}{d x}\cdot x+v=\frac{1+3 v}{1-v}} \\ {\Rightarrow\frac{d v}{d x} \cdot x=\frac{1+3 v}{1-v}-v=\frac{1+3 v-v+v^{2}}{1-v}=\frac{(v+1)^{2}}{1-V}}\end{array}
$$
$$
\therefore \frac{d v}{d x} \cdot x=\frac{(v+1)^{2}}{1-v}
$$
$$
\Rightarrow \frac{1-v}{(v+1)^{2}} \frac{d v}{d x}-\frac{1}{x}=0
$$
This follows the form of $M(v) \frac{d v}{d x}-N(x)=0$

$\therefore$ It's a homogenous equation
$$
\frac{1-v}{(v+1)^{2}} d v=\frac{1}{x} d x
$$
Integrating both sides,
$$
\begin{array}{l}{\quad \int \frac{1-v}{(v+1)^{2}} d v=\int \frac{1}{x} d x} \\
{\Rightarrow \int \frac{-1}{v+1} d v+\int \frac{2}{(v+1)^{2}} d v=\int \frac{1}{x} d x} \\
 {\Rightarrow-\ln |v+1|-2(v+1)^{-1}=\ln |x|+c}
 \\ {\Rightarrow-\ln \left|\frac{y}{x}+1\right|-2\left(\frac{y}{x}+1\right)^{-1}=\ln |x|+c} \\
  {\Rightarrow c-2\left(\frac{y}{x}+1\right)^{-1}=\ln |x \cdot| \frac{y}{x}+1| |} \\
 {\Rightarrow \quad c-2\left(\frac{y}{x}+1\right)^{-1}=\ln |y+x |} \\
  {\Rightarrow \quad e^{c-2\left(\frac{y}{x}+1\right)^{-1}}=y+x} \\
   {\Rightarrow x+y=c e^{-2\left(\frac{y}{x}+1\right)^{-1}}}\end{array}
$$
Title: Re: TUT0801
Post by: XiaolongZhao on September 27, 2019, 02:28:06 PM
Q: t^3 y'+4t^2 y=e^(-t),t<0
    y(-1)=0
    Find the general solution

A:
  Step1: let the coefficient of y' be '1'
             y'+4/t y=e^(-t)/t^3

  Step2: Set u = e^∫〖p(t) dt〗, where p(t) is the coefficient of y
             Set u=e^∫〖4/t dt〗=e^(4ln(-t))=e^(ln〖(-t)^4〗))=e^(ln(t^4))=t^4 (since t<0)

  Step3: Multiply u to both sides
             t^4 y'+4t^3 y=te^(-t)

             (t^4 y)'=te^(-t)

  Step4: Set integral to both sides
            ∫〖(t^4 y)' dt〗=∫〖te^(-t) dt〗

            For ∫〖te^(-t) dt〗, use ‘by parts’ to solve it:
                  u=t,  du=1*dt,
                  dv=e^(-t)dt,   v= -e^(-t)
                 ∫〖te^(-t) dt〗=-te^(-t)-∫〖-e^(-t) dt〗=-te^(-t)-e^(-t)+C

             t^4 y= -te^(-t)-e^(-t)+C

             y= -e^(-t)/t^3 -e^(-t)/t^4 +C/t^4

  Step5: Plug in y(-1)=0

             0=e-e+C, thus, C=0

             Thus, y= -e^(-t)/t^3 -e^(-t)/t^4
Title: Re: TUT0801
Post by: XiaolongZhao on September 27, 2019, 03:07:47 PM
good