InterviewSolution
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InterviewSolution
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हल कीजिए- `(1+y^(2))+ x-e ^(tan^(-1y))(dy)/(dx)=0.` |
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Answer» दिया गया अवकल समीकरण है- `(1+y^(2))+(x-e^(tan^(-1)y))(dy)/(dx) =0` `implies(1+ y^(2))=-(x-e^(tan ^(-1)y))(dy)/(dx)` `implies (dx)/(dy)=(-(x-e^(tan^(-1)y)))/(1+y^(2))` `implies (dx)/(dy) =-(x)/(1+y^(2))+(e^(tan^(-1)y))/(1+y^(2))` `implies(dx)/(dy)+ (1)/(1+y^(2)).x = (e ^(tan^(-1)y))/(1+y^(2))" "...(1)` जो कि x में रैखिक अवकल समीकरण है . समी (1 ) कि तुलना `(dx)/(dy)+Px =Q` से करने पर, `P =(1)/(1+y^(2))` और `Q=(e^(tan^(-1)y))/(1+y^(2))` `thereforeI. F. =e^(intPdy )=e ^(int(1)/(1+y^(2))dy)=e ^(tan ^(-1)y)` अतः अभीष्ट हल है- `x xx(I. F. ) =int Q xx (I. F. )dy +C` `impliesxe ^(tan^(-1)y)=int ((e^(tan^(-1)y))/(1+y^(2)), e^(tan ^(-1)y))dy+C` `impliesxe ^(tan ^(-1)y)=int e ^(t). e^(t) dt +C,` `["माना" tan ^(-1)y=timplies (1)/(1+y^(2))dy = dt]` `impliesxe^(tan ^(-1)y)=int e^(2t) dt+C` `impliesxe^(tan ^(-1)y)= (e ^(2t))/(2)+C` `impliesx.e^(tan ^(-1)y)=1/2 e ^(2 tan^(-1)y)+C.` |
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