y^'=2t,y(0)=1

解答

y^{^{'}} = 2 t, y (0) = 1

y = t^{2} + 1

求解步骤

y^{'} (t) = 2 t

若

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

则

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

y = \int 2 t d t

\int 2 t d t = t^{2} + c_{1}

y = t^{2} + c_{1}

应用初始条件:

y = t^{2} + 1

y = t^{2} + 1

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

y^{^{'}} = \frac{3 x ^{7} e ^{\frac{y}{x}} + x ^{5} y ^{2}}{x ^{6} y}

x \to 2 lim (\frac{x - 7}{x + 3})

\int_{1}^{\infty} \frac{1}{x ^{3} - 1} d x

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