### Author Topic: TUT0401 Quiz 1  (Read 847 times)

#### Yuying Chen

• Jr. Member
• Posts: 14
• Karma: 8
##### TUT0401 Quiz 1
« on: September 27, 2019, 02:00:36 PM »
$\frac{{dy}}{{dx}} = - \frac{{4x + 3y}}{{2x + y}}$

$\frac{{dy}}{{dx}} = - \frac{{4 + \frac{{3y}}{x}}}{{1 + \frac{y}{x}}}$

$u = \frac{y}{x} \Rightarrow y = ux$

$\frac{{dy}}{{dx}} = \frac{{d(ux)}}{{dx}} = \frac{{du}}{{dx}}x + u \cdot 1$

$\frac{{du}}{{dx}}x + u = - \frac{{4 + 3u}}{{2 + u}}$

$\frac{{du}}{{dx}}x = - \frac{{4 + 3u}}{{2 + u}} - \frac{{2u + {u^2}}}{{2 + u}}$

$\frac{{du}}{{dx}}x = \frac{{ - \left( {{u^2} + 5u + 4} \right)}}{{u + 2}}$

$\frac{{du}}{{dx}}x = \frac{{ - (u + 1)(u + 4)}}{{u + 2}}$

$\int {\frac{{u + 2}}{{(u + 1)(u + 4)}}} du = - \int {\frac{1}{x}} dx$

$LHS: \int {\frac{{u + 2}}{{(u + 1)(u + 4)}}} du = \int {\frac{A}{{(u + 1)}}} + \frac{B}{{(u + 4)}}du$

$= \int {\frac{{A(u + 4) + B(u + 1)}}{{(u + 1)(u + 4)}}} du$

$= \int {\frac{{Au + 4A + Bu + B}}{{(u + 1)(u + 4)}}} du$

$= \int {\frac{{(A + B)u + (4A + B)}}{{(u + 1)(u + 4)}}} du$

$\left\{ {\begin{array}{*{20}{l}} {A + B = 1}\\ {4A + B = 2} \end{array} \Rightarrow \left\{ {\begin{array}{*{20}{l}} {A = \frac{1}{3}}\\ {B = \frac{2}{3}} \end{array}} \right.} \right.$

$\therefore \int {\frac{{u + 2}}{{(u + 1)(u + 4)}}} du = \int {\frac{1}{{3(u + 1)}}} + \frac{2}{{3(u + 4)}}du = - \int {\frac{1}{x}} dx$

$\frac{1}{3}\ln |u + 1| + \frac{2}{3}\ln |u + 4| = - \ln |x| + C$

$\ln |u + 1| + \ln |u + 4 {|^2} = - 3\ln |x| + 3C$

$\ln \left| {\frac{y}{x} + 1} \right| + \ln {\left| {\frac{y}{x} + 4} \right|^2} = - 3\ln |x| + 3C$

$\ln \left| {\frac{{y + x}}{x}} \right| + \ln {\left| {\frac{{y + 4x}}{x}} \right|^2} = - 3\ln |x| + 3C$

$\ln |y + x| - \ln |x| + \ln |y + 4x| - 2\ln |x| = - 3\ln |x| + 3C$

$\ln |y + x| + \ln |y + 4x{|^2} = 3C$

${e^{\ln |y + x||y + 4x{|^2}}} = {e^{3C}}$

$|y + x||y + 4x{|^2} = C$
« Last Edit: September 27, 2019, 03:09:29 PM by Yuying Chen »