InterviewSolution
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InterviewSolution
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हल कीजिए- `y-x(dy)/(dx)=a(y^(2)+(dy)/(dx))` |
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Answer» `y-x (dy)/(dx) =a (y^(2)+(dy)/(dx))` `impliesy-x (dy)/(dx) =ay^(2)+a(dy)/(dx)` `implies(y-ay^(2))=(x+a)(dy)/(dx)` `implies (y-ay^(2))dx=(x+a) (dy)/(dx)` `implies (y-ay^(2))dx=(x+a)dy` `implies(dy)/(x+a)=(dy)/(y(1-ay))` समाकलन करने पर, `int (dx)/(x+a)=int (dy)/(y(1- ay))" "...(1)` अब `(1)/(y (1- ay))=A/y+(B)/(1-ay)" "` [आंशिक भिन्न से] `1=A(1-ay)+By" "...(2)` समी (2 ) में, `y =0 ` रखने पर `A =1 ` समी (2 ) में `1-ay=0` अर्थात `y=1/a` रखने पर `B =a ` `therefore (1)/(y(1-ay))=(1)/(y) +(a)/(1-ay)" "...(3)` समी (1 ) और (3 ) से, `int (1)/(1+x)dx=int ((1)/(y) +(a)/(a-ay))dy` ` impliesint (1)/(a+x)dx=int 1/y dy+a int (1)/(1-ay)dy` माना `1-ay =timplies -ady =dt=-(dt)/(a)` `therefore int (1)/(a+x)dx =int 1/y -int(dt)/(t)` `implieslog |a+x|=log y -log |t|+ log C ` `implieslog |a+x|=log |y|-log |1-ay|+ log C` `implieslog |a+x|+log |1-ay|-log |y|=log C` `implieslog |(a+x)(1-ay)|-log | y|=log C` `implieslog ""((1+x)(1-ay))/(|y|)=log C` `implies((a+x)(1-ay))/(y)=C` `implies (x+a)(1-ay)=Cy` |
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