InterviewSolution
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InterviewSolution
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`(x+y)dy+(x-y)dx=0 , y=1` यदि x = 1 |
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Answer» दिया है, `(x+y)dy+(x-y)dx=0` `implies (dy)/(dx)=(y-x)/(x+y)" …(1)"` दिया गया अवकल समीकरण समघातीय है। माना y = vx `implies (dy)/(dx)=v+x(dv)/(dx)` implies समीकरण (2) से, `v+x(dv)/(dx)=(vx-x)/(x+vx)` `impliesv+x(dv)/(dx)=(x(v-1))/(x(1+v))impliesv+x(dv)/(dx)=(v-1)/(1+v)` `impliesx(dv)/(dx)=(v-1)/(1+v)-v` `implies x(dv)/(dx)=(v-1-v-v^(2))/(1+v)` `implies -x(dv)/(dx)=(1+v^(2))/(1+v)=((v+1))/(1+v^(2))dv=-(1)/(x)dx` समाकलन करने पर, `int((v+1))/(1+v^(2))dv=-int(dx)/(x)` `impliesint(v)/(1+v^(2))dv+int(1)/(1+v^(2))dv=-int(dx)/(x)` `implies (1)/(2)log|v^(2)+1|+tan^(-1)v+log|x|=C` `implies (1)/(2)log((y^(2)+x^(2))/(x^(2)))+log|x|+tan^(-1)((y)/(x))=C` `implies log((y^(2)+x^(2))/(x^(2)))+2log|x|+2tan^(-1)((y)/(x))=2C` `implies log((y^(2)+x^(2))/(x^(2)))+logx^(2)+2tan^(-1)((y)/(x))=2C` `implieslog.((y^(2)+x^(2))x^(2))/(x^(2))+2tan^(-1)((y)/(x))=2C` `implies log(x^(2)+y^(2))+2tan^(-1)((y)/(x))=A" " (2C = A " रखने पर")" ....(2)"` दिया है, `x = 1, y = 1` `log2+(2xx(pi)/(4))=A` `implies A=(pi)/(2)+log2` समीकरण (2) में A का मान रखने पर, `log(x^(2)+y^(2))+2tan^(-1)((y)/(x))=(pi)/(2)+log2` जोकि दिए गए अवकल समीकरण का अभीष्ट हल है। |
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