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Question : 79 of 160
Marks:
+1,
-0
Solution:
Consider the equation,
(x3−3xy2)dx=(y3−3x2y)dy = Let
y=vx then
=v+x v+x=| x3−3x(vx)2 |
| (vx)3−3x2vx |
x=−v x= Then,
∫dv=∫ And,
∫dv=∫ ∫dv−3‌∫dv=∫ Let,
1−v4=t −4v3dv=dt v3dv=− And,
v2=m 2vdv=dm vdv= So,
−‌ln‌t−()‌ln‌=ln‌x+ln‌c −‌ln(1−v4)−‌ln() =ln‌x‌c −‌ln(1−)−‌ln() =ln‌x‌c Further simplify the above,
−‌ln[].[]3 =ln‌c‌x ln[[].[]3]− =ln‌c‌x [[].[]3]−=cx c2(x2+y2)2=(y2−x2)
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