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The solution of the primitive integral equation `(x^2+y^2)dy=x ydx`is `y=y(x)dot`If `y(1)=1`and `y(x_0)=e ,`then `x_0`is(a)`( b ) (c)2sqrt(( d ) (e)(( f ) (g) (h) e^(( i )2( j ))( k )-1( l ))( m ))( n ) (o)`(p)(b) `( q ) (r)2sqrt(( s ) (t)(( u ) (v) (w) e^(( x )2( y ))( z )+1( a a ))( b b ))( c c ) (dd)`(ee)(c)`( d ) (e)sqrt(( f )3( g ))( h )e (i)`(j)(d) `( k ) (l)sqrt(( m ) (n) (o)(( p ) (q) e^(( r )2( s ))( t )+1)/( u )2( v ) (w) (x))( y ) (z)`(aa)A. `sqrt(2 (e^(2) - 1))`B. `sqrt(2(e^(2) + 1))`C. `sqrt3 e`D. `sqrt((1)/(2) (e^(2) + 1))` |
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Answer» Correct Answer - C We have `(dy)/(dx) = (xy)/(x^(2) + y^(2))` put y = vx `implies (dy)/(dx) = v .1 + x (dv)/(dx)` `v + x (dv)/(dx) = (v)/(1 +v^(2))` `x(dv)/(dx) = (v)/(1+ v^(2))-v = (v -v-v^(3))/(1+v^(2))` `(1+v^ (2))/(v^(3))dv = -(dx)/(x)` `int (v^(-3) + (1)/(v))dv - int(dx)/(x)` `(v^(-2))/(-2) + "log v " = - "log" x + C ` `-(1)/(2((y)/(x))^(2))+ "log" ((y)/(x)) = - "log"x + C ` `-(x^(2))/(2y^(2)) + "log y" = C " " ....(1)` y(1) = 1 at x = 1 , y = 1 C - 1/2 put in (1) `-(x^(2))/(2y^(2)) + "log y" = -(1)/(2) ` `y(x_(0)) = e`, at x = `x_(0) , y = e` `-(x_(0)^(2))/(2e^(2)) + 1= (-1)/(2)` `x_(0) = sqrt3 e ` |
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