Coefficients linear in two variables
(3x − y + 2) dx + (9x − 3y + 1)dy = 0
(3x−y+2)dx+(9x−3y+1)dy=0u=3x−y+2x=u+y−23dx=13(du+dy)u⋅13(du+dy)+(3u−5)dy=0(10u−5)dy+udu=0y=−∫udu10u−5=−(110∫du+12∫110u−5du)==−110u+120ln∣u−12∣+Cy=−110(3x−y+2)+120ln∣3x−y+32∣+C\left( 3x-y+2 \right) dx+\left( 9x-3y+1 \right) dy=0\\u=3x-y+2\\x=\frac{u+y-2}{3}\\dx=\frac{1}{3}\left( du+dy \right) \\u\cdot \frac{1}{3}\left( du+dy \right) +\left( 3u-5 \right) dy=0\\\left( 10u-5 \right) dy+udu=0\\y=-\int{\frac{udu}{10u-5}}=-\left( \frac{1}{10}\int{du}+\frac{1}{2}\int{\frac{1}{10u-5}du} \right) =\\=-\frac{1}{10}u+\frac{1}{20}\ln \left| u-\frac{1}{2} \right|+C\\y=-\frac{1}{10}\left( 3x-y+2 \right) +\frac{1}{20}\ln \left| 3x-y+\frac{3}{2} \right|+C(3x−y+2)dx+(9x−3y+1)dy=0u=3x−y+2x=3u+y−2dx=31(du+dy)u⋅31(du+dy)+(3u−5)dy=0(10u−5)dy+udu=0y=−∫10u−5udu=−(101∫du+21∫10u−51du)==−101u+201ln∣∣u−21∣∣+Cy=−101(3x−y+2)+201ln∣∣3x−y+23∣∣+C
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