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

解

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

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

解答ステップ

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

線形 ODE を解く:

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

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