y^{''}+5y^'+4y=sin(3x)

解答

y^{^{''}} + 5 y^{^{'}} + 4 y = sin (3 x)

y = c_{1} e^{- x} + c_{2} e^{- 4 x} - \frac{1}{50} sin (3 x) - \frac{3}{50} cos (3 x)

求解步骤

y^{''} (x) + 5 y^{'} (x) + 4 y = sin (3 x)

求解线性方程:

y = c_{1} e^{- x} + c_{2} e^{- 4 x} - \frac{1}{50} sin (3 x) - \frac{3}{50} cos (3 x)

y = c_{1} e^{- x} + c_{2} e^{- 4 x} - \frac{1}{50} sin (3 x) - \frac{3}{50} cos (3 x)

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

9 (x^{2} + y^{2}) d x + 7 x y d y = 0

n = 1 \sum \infty \frac{64}{n ^{3}}

\int \frac{6 x ^{2} - 3 x + 1}{( 4 x + 1 ) ( x ^{2} + 1 )} d x

d er i v a t i v e y = 2^{3 x^{2}}