A curve y=f(x) lying in the first quadrant has the property that the y -intercept of the tangent to the curve at any point (x,y) is equal to - x^2 + y^2 . If the curve passes through the point (3,4) , then the value of y when x=5 is equal to

Step-by-step solution

The equation of the tangent to the curve y=f(x) at a point (x,y) is given by: Y - y = dy dx (X - x) To find the y -intercept, set X = 0 : Y_ int = y - x dy dx According to the given condition: y - x dy dx = - x^2 + y^2 x dy dx - y = x^2 + y^2 This is a homogeneous differential equation. Let y = vx , which implies dy dx = v + x dv dx . Substituting these into the equation gives: x (v + x dv dx ) - vx = x^2 + v^2x^2 x^2 dv dx = x 1 + v^2 (since x > 0 in the first quadrant) dv 1 + v^2 = dx x Integrating both sides: dv