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[tex]1) y''-5y'=x^2-3x\\\\2) y'sinx+4ycosx=\frac{3}{sin^2x}[/tex]
[tex]1) y''-5y'=x^2-3x\\\\2) y'sinx+4ycosx=\frac{3}{sin^2x}[/tex]
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решение на фотографиях