ty^'-2y=t^5sin(2t)-t^3+4t^4

解答

t y^{^{'}} - 2 y = t^{5} sin (2 t) - t^{3} + 4 t^{4}

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

求解步骤

t y^{'} (t) - 2 y = t^{5} sin (2 t) - t^{3} + 4 t^{4}

求解线性方程:

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

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

t an g e n t f (x) = \frac{12 x}{x ^{2} - 4}, at x = 4

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