InterviewSolution
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InterviewSolution
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`2xydx+(x^(2)+2y^(2))dy=0` को हल कीजिए। |
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Answer» दिया है : `2xydx+(x^(2)+2y^(2))dy=0` `implies(dy)/(dx)=-(2xy)/(x^(2)+2y^(2))" "......(1)` `implies(dy)/(dx)=-(2xy)/(x^(2)+2y^(2))" "......(1)` समीकरण (1) में y = vx अर्थात `(dy)/(dx)=-(2x(vx))/(x^(2)+2v^(2)x^(2))=-(2v)/(1+2v^(2))` `impliesx(dv)/(dx)=-(2v)/(1+2v^(2))-v=-((3v+2v^(2)))/(1+2v^(2))` `implies(1+2v^(2))/(3v+2v^(2))dv+(dx)/(x)=0` दोनों ओर का समाकलन करने पर `int(1+2v^(2))/(3v+2v^(3))dv+int(dx)/(x)=0` `implies(1)/(3)int(dt)/(t)+log|x|=logc_(1)` [माना `t=3v+2v^(3)implies(dt)/(3)=(1+2v^(2))dv]` `implies(1)/(3)log|t|+log|x|=logc_(1)` `implies|t|^(1//3).x=c_(1)` `impliestx^(3)=c_(1)^(3)=c` `implies(3v+2v^(3))x^(3)=c` `implies3vx^(3)+2v^(3)x^(3)=c` पुनः `v=(y)/(x)` रखने पर `3x^(2)y+2y^(3)=c` यही दी गयी समीकरण का अभीष्ट हल है। |
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