求微分方程√(1+x^2)*sin(2y)*y'=2x*sin(y)^2+e^(2√(1+x^2))通解

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求微分方程√(1+x^2)*sin(2y)*y'=2x*sin(y)^2+e^(2√(1+x^2))通解
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求微分方程√(1+x^2)*sin(2y)*y'=2x*sin(y)^2+e^(2√(1+x^2))通解
求微分方程√(1+x^2)*sin(2y)*y'=2x*sin(y)^2+e^(2√(1+x^2))通解

求微分方程√(1+x^2)*sin(2y)*y'=2x*sin(y)^2+e^(2√(1+x^2))通解
如果等式右边是(siny)^2而不是sin(y^2),那么可以解,否则不会做.
令根号(1+x^2)=z,dz/dx=x/z,-cos(2y)=g,则dg/dz=dg/dx*dx/dz=2sin(2y)*dy/dx*z/x,代入得
x/2*dg/dz=2x*(1+g)/2+e^(2z),即dg/dz=2g+2+2e^(2z)/根号(z^2-1),于是
(e^(-2z)g)'=e^(-2z)*(g'-2g)=2*e^(-2z)+2/根号(z^2-1),由此得特解为
e^(-2z)g=-e^(-2z)+2ln(z+根号(z^2-1)),g(z)=-1+2e^(2z)ln(z+根号(z^2-1)).因此通解为
g(z)=Ce^(2z)-1+2e^(2z)ln(z+根号(z^2-1)),即-cos2y=Ce^(2根号(1+x^2))-1+2e^(2根号(1+x^2))ln(x+根号(1+x^2)).

√(1+x^2)d(siny)^2=2x(siny)^2 *e^(2√(1+x^2)dx
d(siny)^2=(siny)^2 * e^(2√(1+x^2)) *[2x/√(1+x^2)]dx
d(siny)^2/(siny)^2=e^(2√(1+x^2)d2√(1+x^2)
ln|siny^2|=e^(2√(1+x^2)) +C