y^{''}+3y^'=x^2sin(x)cos(x)

解

y^{^{''}} + 3 y^{^{'}} = x^{2} sin (x) cos (x)

y = c_{1} + c_{2} e^{- 3 x} + \frac{1}{6} (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))) + e^{- 3 x} \cdot \int - \frac{x ^{2} sin ( x ) cos ( x )}{3 e ^{- 3 x}} d x

解答ステップ

y^{''} (x) + 3 y^{'} (x) = x^{2} sin (x) cos (x)

線形 ODE を解く:

y = c_{1} + c_{2} e^{- 3 x} + \frac{1}{6} (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))) + e^{- 3 x} \cdot \int - \frac{x ^{2} sin ( x ) cos ( x )}{3 e ^{- 3 x}} d x

y = c_{1} + c_{2} e^{- 3 x} + \frac{1}{6} (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))) + e^{- 3 x} \cdot \int - \frac{x ^{2} sin ( x ) cos ( x )}{3 e ^{- 3 x}} d x