求下列微分方程的同解1.xy''=y'ln(y'/x) y=1/C1(x-1/C1)e^(C1x+1)+C22.yy''-y'^2=y^2*y' y=C1C2e^(C1x)/(1-C2e^(C1x)) ; y=C2
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![求下列微分方程的同解1.xy''=y'ln(y'/x) y=1/C1(x-1/C1)e^(C1x+1)+C22.yy''-y'^2=y^2*y' y=C1C2e^(C1x)/(1-C2e^(C1x)) ; y=C2](/uploads/image/z/14058815-23-5.jpg?t=%E6%B1%82%E4%B8%8B%E5%88%97%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8B%E7%9A%84%E5%90%8C%E8%A7%A31.xy%27%27%3Dy%27ln%28y%27%2Fx%29+y%3D1%2FC1%28x-1%2FC1%29e%5E%28C1x%2B1%29%2BC22.yy%27%27-y%27%5E2%3Dy%5E2%2Ay%27+y%3DC1C2e%5E%28C1x%29%2F%281-C2e%5E%28C1x%29%29+%3B+y%3DC2)
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