A curve x = x(y) passes through the point (1, 1) and satisfies the differential equation y^5(y^3+1) dx + (3xy^4 - (y^3+1)^3) dy = 0 for y > 0 . The value of x(2) is equal to

Step-by-step solution

The given differential equation can be rewritten as: y^5(y^3+1) dx dy + 3xy^4 = (y^3+1)^3 dx dy + 3 y(y^3+1) x = (y^3+1)^2 y^5 This is a linear differential equation of the form dx dy + P(y)x = Q(y) . Integrating Factor (I.F.) = e^ 3 y(y^3+1) dy = e^ 3y^2 y^3(y^3+1) dy Let y^3 = t 3y^2 dy = dt I.F. = e^ 1 t(t+1) dt = e^ ( 1 t - 1 t+1 ) dt = e^ ( t t+1 ) = y^3 y^3+1 The solution is given by: x ( I.F. ) = Q(y) ( I.F. ) dy x ( y^3 y^3+1 ) = (y^3+1)^2 y^5 y^3 y^3+1 dy x ( y^3 y^3+1 ) = y^3+1 y^2 dy = ( y + 1 y^2 ) dy x