y^{''}+5y=0,y^'(1)=3

Lösung

y^{^{''}} + 5 y = 0, y^{^{'}} (1) = 3

y = - \frac{cos ( 5 t ) ( 3 - 5 cos ( 5 ) c _{2} )}{5 sin ( 5 )} + c_{2} sin (5 t)

Schritte zur Lösung

y^{''} (t) + 5 y = 0

Solve linear ODE:

y = - \frac{cos ( 5 t ) ( 3 - 5 cos ( 5 ) c _{2} )}{5 sin ( 5 )} + c_{2} sin (5 t)

y = - \frac{cos ( 5 t ) ( 3 - 5 cos ( 5 ) c _{2} )}{5 sin ( 5 )} + c_{2} sin (5 t)

x \to 0 + lim (tan (x))

\int \frac{1}{cos ( 2 y ) - cos ^{2} ( y )} d y

t an g e n t f (x) = (x - 2) (x^{2} + 4), (1, - 5)

a re a sin (x), 0, π

4 y^{^{'''}} + 12 y^{^{''}} - 16 y^{^{'}} - 48 y = 0