1求y'=e的(x+Y)次方的通解2求ylnxdx+xlnydy=0的通解

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