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

解答

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

y = \frac{e ^{2 t}}{3} - \frac{4}{3 e ^{t}}

求解步骤

y^{'} (t) + y = e^{2 t}

求解线性方程:

y = \frac{e ^{2 t}}{3} - \frac{4}{3 e ^{t}}

y = \frac{e ^{2 t}}{3} - \frac{4}{3 e ^{t}}

\int (1 - ln (x)) d x

\frac{\partial}{\partial y} (\frac{e ^{x} + e ^{y}}{2})

y^{^{''}} + 2 y^{^{'}} + y = 1, y (1) = 1, y^{^{'}} (1) = 0

x \to - 5 + lim (\frac{( x - 5 )}{x + 5})

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