• عربي

Need Help?

Subscribe to Differential Equation

\${selected_topic_name}
• Notes

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 :

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