y^'+ycos(x)=sin(x)cos(x)

解答

y^{^{'}} + y \cdot cos (x) = sin (x) \cdot cos (x)

y = sin (x) - 1 + \frac{c _{1}}{e ^{s i n (x)}}

求解步骤

y^{'} (x) + y \cdot cos (x) = sin (x) \cdot cos (x)

求解线性方程:

y = sin (x) - 1 + \frac{c _{1}}{e ^{s i n (x)}}

y = sin (x) - 1 + \frac{c _{1}}{e ^{s i n (x)}}

d er i v a t i v e 3 3 x^{2}

\int \frac{e ^{x} + sec ^{2} ( x )}{e ^{x} + tan ( x )} d x

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\int_{0}^{4} x x^{2} + 9 d x

\int \frac{1}{x ^{4} x ^{2} + 3} d x