y+y^4=(y^6+3x)(dy)/(dx)

解答

y + y^{4} = (y^{6} + 3 x) \frac{d y}{d x}

y = - 3 \frac{3 c _{1} - 3 x + ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}, y = - 3 \frac{3 c _{1} - 3 x - ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}

求解步骤

y + y^{4} = (y^{6} + 3 x) \frac{d y}{d x}

求解恰当方程:

y = - 3 \frac{3 c _{1} - 3 x + ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}, y = - 3 \frac{3 c _{1} - 3 x - ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}

y = - 3 \frac{3 c _{1} - 3 x + ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}, y = - 3 \frac{3 c _{1} - 3 x - ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}

d er i v a t i v e y = (\frac{( 3 x ^{2} )}{( 2 x + 4 )})^{3}

x \to 2 + lim (1)

\int x + x^{2} + 2 d x

\frac{d}{d x} (x^{3} + x^{2} y^{3} - 2 y^{2})

\frac{d}{d x} ((4 x^{2} + 1)^{2})