sqrt(1-6x^2)y^'=x

解答

1 - 6 x^{2} y^{^{'}} = x

y = - \frac{1}{6} 1 - 6 x^{2} + c_{1}

求解步骤

1 - 6 x^{2} y^{'} (x) = x

整理

y^{'} (x) : y^{'} (x) = \frac{x}{1 - 6 x ^{2}}

若

f^{^{'}} (x) = g (x)

则

f (x) = \int g (x) d x

y = \int \frac{x}{1 - 6 x ^{2}} d x

\int \frac{x}{1 - 6 x ^{2}} d x = - \frac{1}{6} 1 - 6 x^{2} + c_{1}

y = - \frac{1}{6} 1 - 6 x^{2} + c_{1}

a re a x = - 2, y = x + 2, y = x^{2}

\int 8 y d y

m = 2 \sum \infty \frac{5}{6 ^{m}}

y^{^{'}} = 8 x^{6} y - y, y (1) = - 4

d er i v a t i v e \frac{3 x}{x}