4y^{''}-16y^'+64y=-(2e^{8x})/(3x)

Lösung

4 y^{^{''}} - 16 y^{^{'}} + 64 y = - \frac{2 e ^{8 x}}{3 x}

y = e^{2 x} (c_{1} cos (23 x) + c_{2} sin (23 x)) + e^{2 x} cos (23 x) \cdot \int \frac{3 e ^{6 x} sin ( 2 3 x )}{36 x} d x + e^{2 x} sin (23 x) \cdot \int - \frac{3 e ^{6 x} cos ( 2 3 x )}{36 x} d x

Schritte zur Lösung

4 y^{''} (x) - 16 y^{'} (x) + 64 y = - \frac{2 e ^{8 x}}{3 x}

Solve linear ODE:

y = e^{2 x} (c_{1} cos (23 x) + c_{2} sin (23 x)) + e^{2 x} cos (23 x) \cdot \int \frac{3 e ^{6 x} sin ( 2 3 x )}{36 x} d x + e^{2 x} sin (23 x) \cdot \int - \frac{3 e ^{6 x} cos ( 2 3 x )}{36 x} d x

y = e^{2 x} (c_{1} cos (23 x) + c_{2} sin (23 x)) + e^{2 x} cos (23 x) \cdot \int \frac{3 e ^{6 x} sin ( 2 3 x )}{36 x} d x + e^{2 x} sin (23 x) \cdot \int - \frac{3 e ^{6 x} cos ( 2 3 x )}{36 x} d x

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

y^{^{''}} + 5 y^{^{'}} + 6 y = x^{2}

y^{^{''}} + 9 y = e^{t}

y^{^{''}} + y^{^{'}} - 6 y = x

y^{^{''}} - 7 y^{^{'}} + a = 0, y (0) = 1, y^{^{'}} (0) = 10