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

解

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

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{6 e ^{2} x sin ( x )}{5} - \frac{7 e ^{2} x cos ( x )}{5} + (- 8 e^{2} + \frac{178 e ^{2}}{25}) sin (x) - \frac{54 e ^{2}}{25} cos (x)

解答ステップ

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

線形 ODE を解く:

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{6 e ^{2} x sin ( x )}{5} - \frac{7 e ^{2} x cos ( x )}{5} + (- 8 e^{2} + \frac{178 e ^{2}}{25}) sin (x) - \frac{54 e ^{2}}{25} cos (x)

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{6 e ^{2} x sin ( x )}{5} - \frac{7 e ^{2} x cos ( x )}{5} + (- 8 e^{2} + \frac{178 e ^{2}}{25}) sin (x) - \frac{54 e ^{2}}{25} cos (x)