What is the solution for y^{''}+5y^'+4y=sin(3x) ?

The solution for y^{''}+5y^'+4y=sin(3x) is y=c_{1}e^{-x}+c_{2}e^{-4x}-1/50 sin(3x)-3/50 cos(3x)

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

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)

Solution steps

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

Solve linear ODE:

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

y = 2^{3 x^{2}}

What is the solution for y^{''}+5y^'+4y=sin(3x) ?
The solution for y^{''}+5y^'+4y=sin(3x) is y=c_{1}e^{-x}+c_{2}e^{-4x}-1/50 sin(3x)-3/50 cos(3x)