Question

(2x+y-4)dx+(x-3y+12)dy=0 solve using COEFFICIENT LINEAR IN TWO VARIABLES

Answer 1

$Given\mathrm{:}\ \left(\mathrm{2}x{}+y-\mathrm{4}\right)dx{}{}{}{}\ \ +\ \left(x-\mathrm{3}y+\mathrm{12}\right)dy\ \ =\ \ 0 \\ \\ we\ \ can\ \ rewrite\ \ in\ \ s\mathrm{tan}dard\ \ form\ \ as\ \ \mathrm{:} \\ \\ \left(\mathrm{2}x{}+y-\mathrm{4}\right)dx{}{}{}{}\ \ =\ \left(-x+\mathrm{3}y-\mathrm{12}\right)dy\ \ \\ \\ \frac{dy}{dx}=\frac{\left(\mathrm{2}x+y-\mathrm{4}\right)}{\left(-x+\mathrm{3}y-\mathrm{12}\right)}\ \ \\ \\ For\ \ coefficient\ \ linear\ \ in\ \ two\ \ \mathrm{var}iables \\ \\ For\ \ \ x=X+h\ \ \ \ \ ,\ \ \ y=Y+k\ \ \\ \\ \frac{dY}{dX}=\frac{\mathrm{2}X+\mathrm{2}h+Y+k-\mathrm{4}}{-X-h+\mathrm{3}Y+\mathrm{3}k-\mathrm{12}}\ \ \ \ =\ \ \frac{\mathrm{2}X+Y+\left(\mathrm{2}h+k-\mathrm{4}\right)}{-X+\mathrm{3}Y+\mathrm{(3}k-h-\mathrm{12)}} \\ \\ Let\ \ \ \mathrm{2}h+k-\mathrm{4}=0\ \ ,\ \ \mathrm{3}k-h-\mathrm{12}=0\ \ , \\ \\ Solving\ \ this\ \ \ two\ \ gives\mathrm{:}\ \ h=0,k=\mathrm{4,} \\ \\ \frac{dY}{dX}=\frac{\mathrm{2}X+Y}{-X+\mathrm{3}Y}\ \ \ \ \ \ \left(Now\ \ \ it\ \ is\ \ \mathrm{hom}ogeneous\right) \\ \\ we\ \ can\ \ rewrite\ \ this\ \ as\ \ \\ \left(\mathrm{3}Y-X\right)dY{}{}\ \ =\ \ \left(\mathrm{2}X+Y\right)dX \\ \\ \mathrm{3}YdY\ \ -\ \ XdY{}{}\ \ =\ \ \mathrm{2}XdX\ \ +\ \ YdX \\ \\ \mathrm{3}YdY\ \ -\ \ XdY{}{}\ \ -\ \mathrm{2}XdX\ \ -\ YdX=0 \\ \\ \int{\mathrm{3}YdY}\ \ -\ \int{\mathrm{2}XdX}\ \ -\ d\left(XY\right)=C \\ \frac{\mathrm{3}Y^{\mathrm{2}}}{\mathrm{2}}-X^{\mathrm{2}}-XY=C \\ \\ \mathrm{3}Y^{\mathrm{2}}-\mathrm{2}XY-\mathrm{2}X^{\mathrm{2}}=\mathrm{2}C \\ \\ \mathrm{Re}place\ \ \ X=x\ \ \ \ and\ \ \ Y=y-\mathrm{4} \\ \\ \mathrm{3}{\left(y-\mathrm{4}\right)}^{\mathrm{2}}-\mathrm{2}x\left(y-\mathrm{4}\right)-\mathrm{2}x^{\mathrm{2}}=\mathrm{2}C \\$