2. (у? — 2ху + 6х)dx — (х? — 2ху+ 2)dy %3D 0, whenх%3D 1,у %3D2 = 3. v(2uv? — 3)du + (3и?v? — Зи + 4v)dv %3D 0, when и %3D 1, v %3D 1 - 4. (1+ у? + ху?)ӑх + (x?у + у + 2ху)dy — 0

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