y^'+y/t =t^2y^4

解答

y^{^{'}} + \frac{y}{t} = t^{2} y^{4}

y = 3 \frac{1}{t ^{3} ( - 3 ln ( t ) + c _{1} )}

求解步骤

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

求解贝努力方程:

y = 3 \frac{1}{t ^{3} ( - 3 ln ( t ) + c _{1} )}

y = 3 \frac{1}{t ^{3} ( - 3 ln ( t ) + c _{1} )}

x \to + 4 + lim (\frac{7}{x - 4})

\int \frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} d x

t \frac{d y}{d t} = t^{3} + 15 t^{3} y

- 6 y^{^{'}} + 18 x^{2} = 0

(t^{3}) d x + (x t^{2} + x^{3}) d t = 0