Differential EquationCoefficients Linear In Two Variables Problem 1

Find the particular solution of $$(x+y-2) d x-(x+4 y-2) d y=0$$ when $$y=1, x=2$$

$$x+y-2=0$$
$$x+4 y-2=0$$

$$x=-y+2$$
$$-y+2+4 y-2=0$$

$$3 y=0 \quad y=0$$

$$x=2$$

Intersection point $$p(h, k)=p(2,0)$$

$$x=u+2 \rightarrow d x=d u$$
$$y=w+0 \rightarrow d y=d w$$

$$(u+2+w-2) d u-(u+2+4(w)-2) d w=0$$

$$(u+w) d u-(u+4 w) d w=0$$

$$H \cdot E$$

$$(u+w) d u=(u+4 w) d w$$

$$\frac{(u+w) d u}{d w}=\frac{(u+4 w)}{(u+w)}$$

$$\frac{d u}{d w}=\frac{u+4 w}{u+w} \dots-(1)$$

$$u=v w \quad \frac{d u}{d w}=\frac{d v}{d w} w+v \dots(2)$$

$$(2)=(1) \quad \frac{d v\:w}{d w} +v=\frac{u+4 w}{u+w}$$

$$w \frac{d v}{d w}=\frac{v w+4 w}{v w+w}-v$$

$$w \frac{d v}{d w}=\frac{w(v+4)}{w(v+1)}-v$$

$$\times \frac{(v+1)}{(v+1)}$$

$$w \frac{d v}{d w}=\frac{v+4-v^{2}-v}{v+1}$$

$$d w \quad \frac{d v}{d w}=\frac{v+4-v^{2}-v}{v+1} \cdot d w$$

$$w d v=\frac{-v^{2}+4}{v+1} \cdot d w$$

$$\frac{v+1}{\left(-v^{2}+4\right) w} \cdot w d v=\frac{-v^{2}+4}{v+1} \cdot d w \quad \frac{v+1}{\left(-v^{2}+4\right) w}$$

$$\frac{(v+1) d v}{4-v^{2}}=\frac{d w}{w}$$ integrate

$$-\frac{3}{4} \ln |2-v|-\frac{1}{4} \ln |2+v|=\ln |w|+c$$

$$u=v \cdot w \Rightarrow v=\frac{u}{w}$$

$$x=u+2 \Rightarrow u=x-2$$

$$y=w$$

$$v=\frac{u}{w}=\frac{x-2}{w}=\frac{x-2}{y}$$

$$\frac{a t}{x=2, y=1} \frac{2-2}{1}=0$$

$$v=0$$

$$y=w=1$$

$$v=\frac{x-2}{y}$$

$$-\frac{3}{4} \ln |2|-\frac{1}{4} \ln |{2} |=\ln (1)+c$$

$$\ln (2)\left[\frac{-3}{4}-\frac{1}{4}\right]=c$$

$$c=-\ln (2$$

$$v=0$$

The particular Solution is :

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