高数,第六题,求过程
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(1+x)^2=sec^2 u
sec^4 u(y对x的二次导数)+2tanusec^2 u(y对x的一次导数)+y=sinu
dy/dx=(dy/du)*(du/dx)=(dy/du)/(dx/du)=(dy/du)/sec^2 u
d^2 y/dx^2=d(dy/dx)/dx=[d(dy/dx)/du]/(dx/du)=[d(dy/dx)/du]/sec^2 u=[d((dy/du)/(dx/du))/du]/sec^2 u=[d((dy/du)*cos^2 u)/du]/sec^2 u=[d[(dy/du)/du*cos^2 u])/du]/sec^2 u=d^y/du^cos^2/sec^2 u-dy/du*2sinucosu/sec^2 u=d^2y/du^2/sec^4 u-2tanudy/du/sec^4 u
代入原式 d^2y/du^2+y=sinu
sec^4 u(y对x的二次导数)+2tanusec^2 u(y对x的一次导数)+y=sinu
dy/dx=(dy/du)*(du/dx)=(dy/du)/(dx/du)=(dy/du)/sec^2 u
d^2 y/dx^2=d(dy/dx)/dx=[d(dy/dx)/du]/(dx/du)=[d(dy/dx)/du]/sec^2 u=[d((dy/du)/(dx/du))/du]/sec^2 u=[d((dy/du)*cos^2 u)/du]/sec^2 u=[d[(dy/du)/du*cos^2 u])/du]/sec^2 u=d^y/du^cos^2/sec^2 u-dy/du*2sinucosu/sec^2 u=d^2y/du^2/sec^4 u-2tanudy/du/sec^4 u
代入原式 d^2y/du^2+y=sinu
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