ty^'+2y=sin(t),y(pi/2)=3

Lösung

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 3

y = - \frac{cos ( t )}{t} + \frac{sin ( t )}{t ^{2}} + \frac{3 π ^{2} - 4}{4 t ^{2}}

Schritte zur Lösung

t y^{'} (t) + 2 y = sin (t)

Solve linear ODE:

y = - \frac{cos ( t )}{t} + \frac{sin ( t )}{t ^{2}} + \frac{3 π ^{2} - 4}{4 t ^{2}}

y = - \frac{cos ( t )}{t} + \frac{sin ( t )}{t ^{2}} + \frac{3 π ^{2} - 4}{4 t ^{2}}

2 x^{2} y^{^{''}} + 5 x y^{^{'}} + y = 18 x + 4

y^{^{'}} + y^{2} sin (x) = 0

\frac{d y}{d x} = \frac{2 y}{2 x + 1}

t an g e n t 10 x^{2} + 2 x

\int 8 t^{2} d t