The slope of the tangent to a curve C: y=y(x) at any point (x,y) is given by y x + x y ^2 ( y x ) . If the curve passes through the point (1, 2 ) and intersects the line y = 4 x at the point ( , ) , then the value of (2 ^2) is equal to

Step-by-step solution

The given differential equation is: dy dx = y x + x y ^2 ( y x ) This is a homogeneous differential equation. Substitute y = vx , which gives dy dx = v + x dv dx . Substituting this into the equation: v + x dv dx = v + 1 v ^2 v x dv dx = ^2 v v v ^2 v , dv = dx x Integrating the left side by parts (taking v as the first function and ^2 v as the second): -v v - (- v) , dv = |x| + C -v v + | v| = |x| + C The curve passes through (1, 2 ) , so when x = 1 , v = y x = 2 . - 2 ( 2 ) + | ( 2 ) | = (1) + C 0 + 0 = 0 + C C =