y^{''}+9y=sin^4(x)

解

y^{^{''}} + 9 y = sin^{4} (x)

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

解答ステップ

y^{''} (x) + 9 y = sin^{4} (x)

線形 ODE を解く:

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

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