求x*(dy/dx)=lnx-y的通解
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求x*(dy/dx)=lnx-y的通解
![求x*(dy/dx)=lnx-y的通解](/uploads/image/z/3391633-1-3.jpg?t=%E6%B1%82x%2A%28dy%2Fdx%29%3Dlnx-y%E7%9A%84%E9%80%9A%E8%A7%A3)
x*(dy/dx)=lnx-y
y'+1/x*y=(lnx)/x
y' + p(x)•y = q(x)的通解为:
y = [e^-∫ p(x) dx] • [∫ q(x)•[e^∫ p(x) dx] dx + C]
本题
p(x)=1/x
q(x)=lnx/x
∫ p(x) dx=lnx
∫ q(x)•[e^∫ p(x) dx] dx
=∫lnx/x*e^(lnx)dx
=∫lnxdx
=xlnx-∫x*1/xdx
=xlnx-x
故
y = [e^-∫ p(x) dx] • [∫ q(x)•[e^∫ p(x) dx] dx + C]
=e^(-lnx)• (xlnx-x+ C)
=lnx-1+C/x
y'+1/x*y=(lnx)/x
y' + p(x)•y = q(x)的通解为:
y = [e^-∫ p(x) dx] • [∫ q(x)•[e^∫ p(x) dx] dx + C]
本题
p(x)=1/x
q(x)=lnx/x
∫ p(x) dx=lnx
∫ q(x)•[e^∫ p(x) dx] dx
=∫lnx/x*e^(lnx)dx
=∫lnxdx
=xlnx-∫x*1/xdx
=xlnx-x
故
y = [e^-∫ p(x) dx] • [∫ q(x)•[e^∫ p(x) dx] dx + C]
=e^(-lnx)• (xlnx-x+ C)
=lnx-1+C/x