(t^2+16)^2y^'+2t(t^2+16)y=4,y(-2)=10

解

(t^{2} + 16)^{2} y^{^{'}} + 2 t (t^{2} + 16) y = 4, y (- 2) = 10

y = \frac{arctan ( \frac{t}{4} )}{t ^{2} + 16} + \frac{200 + arctan ( \frac{1}{2} )}{t ^{2} + 16}

解答ステップ

(t^{2} + 16)^{2} y^{'} (t) + 2 t (t^{2} + 16) y = 4

線形 ODE を解く:

y = \frac{arctan ( \frac{t}{4} )}{t ^{2} + 16} + \frac{200 + arctan ( \frac{1}{2} )}{t ^{2} + 16}

y = \frac{arctan ( \frac{t}{4} )}{t ^{2} + 16} + \frac{200 + arctan ( \frac{1}{2} )}{t ^{2} + 16}