y^{''}-2y^'+5y=2sin^2(x)

Lösung

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 sin^{2} (x)

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

Schritte zur Lösung

y^{''} (x) - 2 y^{'} (x) + 5 y = 2 sin^{2} (x)

Solve linear ODE:

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

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

so l v e f or d, d (t) = 1399 (0.95)^{t}

y^{^{''}} + (\frac{1}{r}) y^{^{'}} = b

y^{^{''}} - 10 y^{^{'}} + 61 = 0

\frac{d ^{2} x}{d t ^{2}} + 9 \frac{d x}{d t} + 8 x = 2 t

y^{^{''}} - 7 y^{^{'}} + 14 y = - 2 e^{3 t}, y (0) = - 6