ty^'=2y-3t^3cos(4t)

解答

t y^{^{'}} = 2 y - 3 t^{3} cos (4 t)

y = - \frac{3}{4} sin (4 t) t^{2} + c_{1} t^{2}

求解步骤

t y^{'} (t) = 2 y - 3 t^{3} cos (4 t)

求解线性方程:

y = - \frac{3}{4} sin (4 t) t^{2} + c_{1} t^{2}

y = - \frac{3}{4} sin (4 t) t^{2} + c_{1} t^{2}

\frac{\partial}{\partial x} (6 x y - 6 x)

x^{2} y^{^{''}} - 3 x y^{^{'}} - 2 y = 0

x \to - \frac{π}{3} lim (- sec (x))

\int \frac{36 - 25 x ^{2}}{x} d x

\int_{0}^{1} \frac{1}{t ^{2} + 1} d t