(d^2-4d+4)y=e^2x+sin^2(x)

Lösung

(d^{2} - 4 d + 4) y = e^{2} x + sin^{2} (x)

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

Schritte zur Lösung

(d^{2} - 4 d + 4) y = e^{2} x + sin^{2} (x)

Solve linear ODE:

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

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

(d^{2} - 1) \cdot y = x \cdot sinh (x)

y^{^{''}} + y = 12 x

y^{^{''}} + 4 y = 2, e^{x} y (0) = 2, y^{^{'}} (0) = 1

(d^{2} - d - 2) y = 10 cos (x)

y^{^{''}} + 4 y^{^{'}} + 3 y = 4, e^{3} t y (0) = 2, y^{^{'}} (0) = 7