(dx)/(dt)=x^2+1/16 ,x(0)=6

解答

\frac{d x}{d t} = x^{2} + \frac{1}{16}, x (0) = 6

x = \frac{tan ( \frac{t + 4 \cdot 1.52915 \dots}{4} )}{4}

求解步骤

\frac{d x}{d t} = x^{2} + \frac{1}{16}

求解可分离方程:

x = \frac{tan ( \frac{t + 4 \cdot 1.52915 \dots}{4} )}{4}

x = \frac{tan ( \frac{t + 4 \cdot 1.52915 \dots}{4} )}{4}

\frac{d v}{d x} = 9 + 16 v^{2}

y^{^{'}} + \frac{2 x y ^{2} + 2 y}{2 x ^{2} y + 2 x} = 0

\frac{d x}{d y} = \frac{1 + x ^{2}}{1 - y ^{2}}

u^{^{'}} + n^{2} H (t) = - \frac{( - 1 ) ^{n} + 1}{π ( n ^{2} - 1 )}

y^{^{'}} - sin (x) \cdot y = 3 sin (x)