Given the differential equation (sin(x²*y) + 2*x²*y*cos(x²*y)) dx + (x³*cos(x²*y) + 1) dy = 0 with initial conditions y(0) = 0 and y(alpha) = 1, find the value of |alpha * sin(alpha²)|.

Question

Accepted Answer

1. Notice: partialₓ (x³*sin(x²*y)) = 3x²*sin(x²*y) + x³*cos(x²*y)*2xy = 3x²*sin(x²*y) + 2x⁴*y*cos(x²*y). This isn't quite matching. Try: M = sin(x²*y) + 2*x²*y*cos(x²*y), N = x³*cos(x²*y) + 1. Check exactness: dM/dy = x²*cos(x²*y) + 2*x²*cos(x²*y) + 2*x²*y*(-sin(x²*y)*x²) = 3*x²*cos(x²*y) - 2*x⁴*y*sin(x²*y). dN/dx = 3*x²*cos(x²*y) + x³*(-sin(x²*y)*2xy) = 3*x²*cos(x²*y) - 2*x⁴*y*sin(x²*y). Since dM/dy = dN/dx, the equation IS exact. F such that dF/dx = M and dF/dy = N. From dF/dy = x³*cos(x²*y)+1: F = x³*sin(x²*y)/x²... integrate: F = (1/x²)*sin(x²*y)*x³ + y = x*sin(x²*y) + y. Check: dF/dx = sin(x²*y) + x*cos(x²*y)*2xy = sin(x²*y) + 2x²*y*cos(x²*y) = M. YES! dF/dy = x*cos(x²*y)*x² + 1 = x³*co

Given the differential equation (sin(x²y) + 2x²ycos(x²y)) dx + (x³cos(x²y) + 1) dy = 0 with initial conditions y(0) = 0 and y(alpha) = 1, find the value of |alpha sin(alpha²)|.

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