y'''+4y"+4y'=2 solution of non-homogenous differential equation

Using the Method of Undetermined Coefficients, the particular solution is of the form $Ax$.

Note that $y_p(x)=Ax\implies y'_p(x)=A\implies y''_p(x)=0\implies y'''_x(p)=0$

Now, substituting into our original differential equation, we get that $$0+4\cdot 0+4A=2$$ $$\therefore A=\frac 12\implies y_p(x)=\frac x2$$

$$\therefore y=\frac x2+c_1 + c_2e^{-2x}+c_3xe^{-2x}$$

Let $w=y'$, so that the differential equation becomes $$w''+4w'+4w=2.$$

Then let $w_p=A$, and work with $$w_p''+4w_p'+4w_p=2$$

Solve for $w_p$, then integrate to find $y_p$ (since $w_p=y_p'$).

HINT:

$$4y'(x)+4y''(x)+y'''(x)=2\Longleftrightarrow$$

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The general solution will be the sum of the complementary solution and particular solution.

Find the complementary solution by solving:

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$$4y'(x)+4y''(x)+y'''(x)=0\Longleftrightarrow$$

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Assume a solution will be proportional to $e^{\lambda x}$ for some constant $\lambda$.

Substitute $y(x)=e^{\lambda x}$ into the differential equation:

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$$\frac{\text{d}^3}{\text{d}x^3}\left(e^{\lambda x}\right)+4\frac{\text{d}^2}{\text{d}x^2}\left(e^{\lambda x}\right)+4\frac{\text{d}}{\text{d}x}\left(e^{\lambda x}\right)=0\Longleftrightarrow$$

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Substitute $\frac{\text{d}^3}{\text{d}x^3}\left(e^{\lambda x}\right)=\lambda^3e^{\lambda x}$:

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$$\lambda^3e^{\lambda x}+4\lambda^2e^{\lambda x}+4\lambda e^{\lambda x}=0\Longleftrightarrow$$ $$e^{\lambda x}\left(\lambda^3+4\lambda^2+4\lambda\right)=0\Longleftrightarrow$$

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Since $e^{\lambda x}\ne0$ for any finite $\lambda$, the zeros must come from the polynomial:

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$$\lambda^3+4\lambda^2+4\lambda=0\Longleftrightarrow$$ $$\lambda\left(\lambda+2\right)^2=0$$