微分方程,第二题用什么方法求
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微分方程,第二题用什么方法求
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![微分方程,第二题用什么方法求](/uploads/image/z/18963085-13-5.jpg?t=%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8B%2C%E7%AC%AC%E4%BA%8C%E9%A2%98%E7%94%A8%E4%BB%80%E4%B9%88%E6%96%B9%E6%B3%95%E6%B1%82)
∵dy/dx+e^(y^3+x)/y^2=0
==>dy/dx=-e^(y^3+x)/y^2
==>dy/dx=-e^(y^3)*e^x/y^2
==>-3y^2e^(-y^3)dy=3e^xdx
==>e^(-y^3)d(-y^3)=3d(e^x)
==>d(e^(-y^3))=3d(e^x)
==>e^(-y^3)=3e^x+C (C是常数)
==>3e^(y^3+x)+Ce^(y^3)=1
∴原方程的通解是3e^(y^3+x)+Ce^(y^3)=1.
∵y(0)=0
∴代入通解,得3+C=1
==>C=-2
故原方程满足所给初始条件的特解是3e^(y^3+x)-2e^(y^3)=1.
==>dy/dx=-e^(y^3+x)/y^2
==>dy/dx=-e^(y^3)*e^x/y^2
==>-3y^2e^(-y^3)dy=3e^xdx
==>e^(-y^3)d(-y^3)=3d(e^x)
==>d(e^(-y^3))=3d(e^x)
==>e^(-y^3)=3e^x+C (C是常数)
==>3e^(y^3+x)+Ce^(y^3)=1
∴原方程的通解是3e^(y^3+x)+Ce^(y^3)=1.
∵y(0)=0
∴代入通解,得3+C=1
==>C=-2
故原方程满足所给初始条件的特解是3e^(y^3+x)-2e^(y^3)=1.