Is there a simpler method to this solve this equation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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








  • $\begingroup$ Do you see any defects in my answer? I got -2 because of this answer and don't know why?? Thanks for your time. + $\endgroup$ – Mikasa Jan 4 '13 at 6:37
  • $\begingroup$ 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. $\endgroup$ – Elements in Space Jan 4 '13 at 6:55

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: $$X\to x-\alpha\\Y\to y-\beta$$


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