xy^'-y= 3/2 xln(x)

解答

x y^{^{'}} - y = \frac{3}{2} x ln (x)

y = \frac{3}{4} ln^{2} (x) x + c_{1} x

求解步骤

x y^{'} (x) - y = \frac{3}{2} x ln (x)

求解线性方程:

y = \frac{3}{4} ln^{2} (x) x + c_{1} x

y = \frac{3}{4} ln^{2} (x) x + c_{1} x

\int \frac{5 x - 1}{x ^{2} - 3 x - 40} d x

t an g e n t f (x) = \frac{x}{8 - x}, at x = 3

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\int \frac{1}{( 6 + x ) ^{3}} d x

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