What is the solution for y^{''}+4y^'+3y=3x+7 ?

The solution for y^{''}+4y^'+3y=3x+7 is y=c_{1}e^{-x}+c_{2}e^{-3x}+x+1

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y^{''}+4y^'+3y=3x+7

y^{^{''}} + 4 y^{^{'}} + 3 y = 3 x + 7

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + x + 1

Solution steps

y^{''} (x) + 4 y^{'} (x) + 3 y = 3 x + 7

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + x + 1

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + x + 1

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(cos (6 x)) y^{^{'}} + 6 (tan (6 x)) y = - sec (6 x)

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

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t d t + y e^{- t} d y = 0

What is the solution for y^{''}+4y^'+3y=3x+7 ?
The solution for y^{''}+4y^'+3y=3x+7 is y=c_{1}e^{-x}+c_{2}e^{-3x}+x+1