Solve :- y²dx+(xy+x²)dy=0
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Use a substitution:⎧⎩⎨⎪⎪u=yxy=uxdy=udx+xdu
(x2+y2)dx+2xydy=0
(x2+(ux)2)dx+2x(ux)(udx+xdu)=0
(x2+u2x2+2u2x2)dx+2ux3du
x2(1+3u2)dx=−2ux3du
1xdx=−2u1+3u2du
∫1xdx=∫−2u1+3u2du
ln|x|=−13ln|1+3u2|+C0
3ln|x|=−ln|1+3u2|+C1
ln|x3|+ln∣∣1+3⋅y2x2∣∣=C1
ln|x3+3xy2|=C1
x3+3xy2=C
(x2+y2)dx+2xydy=0
(x2+(ux)2)dx+2x(ux)(udx+xdu)=0
(x2+u2x2+2u2x2)dx+2ux3du
x2(1+3u2)dx=−2ux3du
1xdx=−2u1+3u2du
∫1xdx=∫−2u1+3u2du
ln|x|=−13ln|1+3u2|+C0
3ln|x|=−ln|1+3u2|+C1
ln|x3|+ln∣∣1+3⋅y2x2∣∣=C1
ln|x3+3xy2|=C1
x3+3xy2=C
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