3
$\begingroup$

Is there a simpler method to this solve this equation


$(x-2y+1)\text dx+(4x-3y-6)\text dy=0$


$\frac{\text dy}{\text dx}=\frac{2y-x-1}{4x-3y-6}$

$\frac{\text d Y}{ \text d X}=\frac{2(Y+k)-(X+h)-1}{4(X+h)-3(Y+k)-6}$

$\frac{\text d Y}{ \text d X}=\frac{2Y-X+(2k-h-1)}{4X-3Y+(4h-3k-6)}$


$2k-h-1=0\qquad4h-3k-6=0$ $h=2k-1\qquad 4(2k-1)-3k-6=0$ $5k-10=0\qquad k=2\qquad h=2(2)-1\qquad h=3$ $y=Y+2\qquad\qquad x=X+3$ $Y=y-2\qquad\qquad X=x-3$


$\frac{\text d Y}{\text d X}=\frac{2Y-X}{4X-3Y}=\frac{2\left(\frac{Y}{X}\right)-1}{4-3\left(\frac{Y}{X}\right)}$

$\left(\frac{Y}{X}\right)=V\qquad Y=VX$

$\frac{\text d Y}{\text d X}=X\frac{\text d V}{\text d X}+V$

$X\frac{\text d V}{\text d X}+V=\frac{2V-1}{4-3V}$

$X\frac{\text d V}{\text d X}=\frac{2V-1}{4-3V}-V$

$X\frac{\text d V}{\text d X}=\frac{2V-1}{4-3V}-V\left(\frac{4-3V}{4-3V}\right)$

$X\frac{\text d V}{\text d X}=\frac{3V^2-2V-1}{4-3V}$

$\frac{\text dX}{X}=\frac{4-3V}{3V^2-2V-1}\text dV$

$\int\frac{\text dX}{X}=\int\frac{4-3V}{3V^2-2V-1}\text dV$

$\ln X=\int\frac{4-3V}{3V^2-2V-1}\text dV$


$3V^2-2V-1=(V-1) \overline {\bigg)3V^2-2V-1}$

$\qquad3V+1$$=(V-1) \overline {\bigg)3V^2-2V-1}$$\quad\qquad3V^2-3V$$\qquad\qquad\qquad\qquad V-1$$\qquad\qquad\qquad\qquad \underline{V-1}$$\qquad\qquad\qquad\qquad 0$

$3V^2-2V-1=(V-1)(3V+1)$

$\frac{4-3V}{3V^2-2V-1}=\frac{4-3V}{(V-1)(3V+1)}=\frac{\alpha}{(V-1)}+\frac{\beta }{(3V+1)}=\frac{(3V+1)\alpha+(V-1)\beta}{(V-1)(3V+1)}$

$4-3V=(3V+1)\alpha+(V-1)\beta$ $4-3V=3V\alpha+\alpha +V\beta-\beta$ $4-3V=\alpha-\beta+(3\alpha +\beta)V$ $4=\alpha-\beta$ $-3=3\alpha+\beta$ $4+\beta=\alpha$ $-3=3(4+\beta)+\beta$ $-15=4\beta$ $\beta=\frac{-15}{4}$ $4=\alpha-\left(\frac{-15}{4}\right)$ $\alpha=4-\frac{15}{4}=\frac{16}{4}-\frac{15}{4}=\frac14$


$\ln X=\int\frac{1/4}{V-1}-\frac{15/4}{3V+1}\text d V$

$\ln X=\frac{1}{4}\int\frac{1}{V-1}-\frac{15}{3V+1}\text d V$

$4\ln X=\ln(V-1)-5\ln(3V+1)+c$

$\ln \left(X^4\right)=\ln\left({\frac{(V-1)}{(3V+1)^5}}\right)+c$

$X^4=\frac{V-1}{(3V+1)^5}\times e^c$

$(3V+1)^5X^4=e^c(V-1)$

$(3V+1)^5X^5=AX(V-1)$

$(3XV+X)^5=A(XV-X)$

$(3X\tfrac{Y}{X}+X)^5=A(X\tfrac{Y}{X}-X)$

$(3Y+X)^5=A(Y-X)$

$(3(y-2)+(x-3))^5=A((y-2)-(x-3))$

$(3y+x-9)^5=A(y-x+1)$

  • 0
    As far as I can tell your response hasen't added anything, you have outlined the method I have used above. I am looking for a different method that could potential simplify the calculation.2013-01-04

1 Answers 1

1

I just know that when we have a first differential equation as $f(a_1x+b_1y+c_1)dx+g(a_2x+b_2y+c_2)dy=0$ then if two lines $a_1x+b_1y+c_1=0\\a_2x+b_2y+c_2=0 \;\;^*$ are not parallel $(\frac{a_1}{a_2}\neq\frac{b_1}{b_2})$ so we can use the new change of variable as you also did above $x=X+\alpha\\y=Y+\beta$ in which $(\alpha,\beta)$ is the solution of above system of equations$^{*}$. I think you can find the point of intersection of lines first and then omit the constants in your equation (for example +1 and -6) temporary. After solving the homogeneous equation, as you did above, do as following: $X\to x-\alpha\\Y\to y-\beta$