InterviewSolution
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InterviewSolution
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`x^(2)dy+(xy+y^(2))dx=0 , y=1` यदि x = 1 |
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Answer» `x^(2)dy+(xy+y^(2))dx=0` `implies (dy)/(dx)+(xy+y^(2))/(x^(2))=0 " ...(1)"` माना y = vx `implies (dy)/(dx)=v+(xdv)/(dx)` समीकरण (1) से, `v+x(dv)/(dx)+(vx^(2)+v^(2)x^(2))/(x^(2))=0` `implies x(dv)/(dx)+2v+v^(2)=0` `implies x(dv)/(dx)=-v(v+2)` `implies (dv)/(v(v+2))=-(dx)/(x)` `implies int(2)/(v(v+2))dv=-int(2)/(x)dx` `implies int((1)/(v)-(1)/(v+2))dv=-int(2)/(x)dx` `implies logv-log(v+2)=-2logx+logc` `implies log.(v)/(v+2)=log.(c)/(x^(2))` `implies (v)/(v+2)=(c)/(x^(2))` `implies (y//x)/((y)/(x)+2)=(c)/(x^(2))` `implies (y)/(2x+y)=(c)/(x^(2))` `implies x^(2)y=c(2x+y)` दिया है x = 1 पर y = 1 `therefore 1 = c(2+1)` `implies c=(1)/(3)` अतः `x^(2)y=(1)/(3)(2x+y)` `implies 3x^(2)y=2x+y` |
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