y^'=(t+y)^2-1,y(3)=6

解答

y^{^{'}} = (t + y)^{2} - 1, y (3) = 6

y = - \frac{9}{9 t - 28} - t

求解步骤

y^{'} (t) = (t + y)^{2} - 1

使用替代法求解:

y = - \frac{9}{9 t - 28} - t

y = - \frac{9}{9 t - 28} - t

\int \frac{( x ^{2} + 3 x - 4 )}{x ^{3} - 4 x ^{2} + 4 x} d x

a re a y = 4 x^{2}, x = 1, y = 0

in v erse l a pl a ce \frac{2 e ^{- 2 s}}{5 s ^{2} + 45}

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

d er i v a t i v e \frac{x ^{63} - 1}{x - 1}