Question

1 求y'=e的(x+Y)次方的通解 2 求ylnxdx+xlnydy=0的通解

Answer 1

1

dy/dx = e^(x+y)

所以d(x+y)/dx - 1 = e^(x+y)

设t = x+y那么dt/dx = e^t+1

所以dt/(e^t+1) = dx

积分得ln(e^(x+y)/[e^(x+y)-1]) = x+C

2

(lnx)/x dx + (lny) / y dy =0

积分得(lnx)^2+(lny)^2 = C