InterviewSolution
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InterviewSolution
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`(1+e^(x//y))dx+e^(x//y)(1-(x)/(y))dy=0` |
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Answer» `(1+e^(x//y))dx+e^(x//y)(1-(x)/(y))dy=0` `implies (1+e^(x//y))dx=-e^(x//y)(1-(x)/(y))dy` `implies (dx)/(dy)=(-e^(x//y)(1-(x)/(y)))/(1+e^(x//y))" ....(1)"` यह एक समघातीय अवकल समीकरण है। माना x = vy `implies (dx)/(dy)=v+y.(dv)/(dy)` समीकरण (1) में रखने पर, `v+y.(dv)/(dy)=(-e^(v)(1-v))/(1+e^(v))` `implies y(dv)/(dy)=(-e^(v)+ve^(v))/(1+e^(v))-v` `implies y.(dv)/(dy)=(-e^(v)+ve^(v)-v-ve^(v))/(1+e^(v))` `=(-(e^(v)+v))/(1+e^(v))` `implies (1+e^(v))/(v+e^(v))dv=-(dy)/(y)` `implies int(1+e^(v))/(v+e^(v))dv=-int(dy)/(y)` `implies log(v+e^(v))=-logy+logc` `implies log(v+e^(v))+logy=logc` `implies log(vy+ye^(v))=logc` `implies vy+ye^(v)=c` `implies x+y.e^(x//y)=c` |
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