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求一道高等数学微分方程题,图上第四题(不是第4题),求详细过程,不会的勿扰,谢谢
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解:∵ y'=(2x^3+3xy^2-7x)/(3x^2y+2y^3-8y)
==>dy/dx=x(2x^2+3y^2-7)/(y(3x^2+2y^2-8))
==>(2ydy)/(2xdx)=(2x^2+3y^2-7)/(3x^2+2y^2-8)
==>d(y^2)/d(x^2)=(2x^2+3y^2-7)/(3x^2+2y^2-8)
==>dn/dm=(2m+3n)/(3m+2n) (令m=x^2-2,n=y^2-1)
==>(mdt+tdm)/dm=(2+3t)/(3+2t) (令n=mt)
==>m(2t+3)dt+2(t^2-1)dm=0 (化简)
==>(2t+3)dt/(t^2-1)+2dm/m=0
==>[5/(t-1)-1/(t+1)]dt+4dm/m=0
==>∫[5/(t-1)-1/(t+1)]dt+4∫dm/m=0 (积分)
==>5ln│t-1│-ln│t+1│+4ln│m│=ln│C│ (C是非零常数)
==>m^4(t-1)^5/(t+1)=C
==>m^4(n/m-1)^5/(n/m+1)=C
==>(n-m)^5/(n+m)=C
==>(n-m)^5=C(n+m)
==>(y^2-x^2+1)^5=C(y^2+x^2-3)
∴ 此方程的通解是(y^2-x^2+1)^5=C(y^2+x^2-3)。
==>dy/dx=x(2x^2+3y^2-7)/(y(3x^2+2y^2-8))
==>(2ydy)/(2xdx)=(2x^2+3y^2-7)/(3x^2+2y^2-8)
==>d(y^2)/d(x^2)=(2x^2+3y^2-7)/(3x^2+2y^2-8)
==>dn/dm=(2m+3n)/(3m+2n) (令m=x^2-2,n=y^2-1)
==>(mdt+tdm)/dm=(2+3t)/(3+2t) (令n=mt)
==>m(2t+3)dt+2(t^2-1)dm=0 (化简)
==>(2t+3)dt/(t^2-1)+2dm/m=0
==>[5/(t-1)-1/(t+1)]dt+4dm/m=0
==>∫[5/(t-1)-1/(t+1)]dt+4∫dm/m=0 (积分)
==>5ln│t-1│-ln│t+1│+4ln│m│=ln│C│ (C是非零常数)
==>m^4(t-1)^5/(t+1)=C
==>m^4(n/m-1)^5/(n/m+1)=C
==>(n-m)^5/(n+m)=C
==>(n-m)^5=C(n+m)
==>(y^2-x^2+1)^5=C(y^2+x^2-3)
∴ 此方程的通解是(y^2-x^2+1)^5=C(y^2+x^2-3)。
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