y^{''}-2y^'+2y=2x-2,y^'(pi)=pi,y(0)=0

Lösung

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x - 2, y^{^{'}} (π) = π, y (0) = 0

y = e^{x} (- \frac{( π - 1 - ( π - 1 ) ) cos ( x )}{e ^{π}} - \frac{π - 1}{e ^{π}} sin (x)) + x

Schritte zur Lösung

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

Solve linear ODE:

y = e^{x} (- \frac{( π - 1 - ( π - 1 ) ) cos ( x )}{e ^{π}} - \frac{π - 1}{e ^{π}} sin (x)) + x

y = e^{x} (- \frac{( π - 1 - ( π - 1 ) ) cos ( x )}{e ^{π}} - \frac{π - 1}{e ^{π}} sin (x)) + x

\frac{d ^{2} y}{d x ^{2}} + 4 y = sin^{2} (x)

e^{x} y^{^{''}} + 2 e^{x} y^{^{'}} + y e^{x} = ln (x)

y^{^{''}} + 7 y^{^{'}} + 6 y = 24

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

(d^{2} + 6 d + 10) y = 3 x e^{- 3 x} - 2 e^{3 x} cos (x)