y^{''}+9y=sec(3x^{18}x)-36

Lösung

y^{^{''}} + 9 y = sec (3 x^{18} x) - 36

y = c_{1} cos (3 x) + c_{2} sin (3 x) + cos (3 x) \cdot \int - \frac{sin ( 3 x ) ( sec ( 3 x ^{19} ) - 36 )}{3} d x + sin (3 x) \cdot \int \frac{cos ( 3 x ) ( sec ( 3 x ^{19} ) - 36 )}{3} d x

Schritte zur Lösung

y^{''} (x) + 9 y = sec (3 x^{18} x) - 36

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + cos (3 x) \cdot \int - \frac{sin ( 3 x ) ( sec ( 3 x ^{19} ) - 36 )}{3} d x + sin (3 x) \cdot \int \frac{cos ( 3 x ) ( sec ( 3 x ^{19} ) - 36 )}{3} d x

y = c_{1} cos (3 x) + c_{2} sin (3 x) + cos (3 x) \cdot \int - \frac{sin ( 3 x ) ( sec ( 3 x ^{19} ) - 36 )}{3} d x + sin (3 x) \cdot \int \frac{cos ( 3 x ) ( sec ( 3 x ^{19} ) - 36 )}{3} d x

\int 5^{t} d t

\int \frac{1}{u ^{2} + xu} d u

\int_{0}^{1} 1 - 2 x - 3 x^{2} d x

\frac{d y}{d x} y = (4 x - 7)^{2}

s l o p e (7, 5), (3, 2)