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This is non-homogenous differential equation : $$y'''+4y''+4y'=2.$$

Of course, I started with characteristic polynomial of homogenous case:

$$t^3+4t^2+4t=0$$ then $$t(t^2+4t+4)=0$$ we have: $$t_1=0; t_{2,3}=-2.$$ So, solution of homogenous case is:

$$y_s(x)=c_1 + c_2e^{-2x}+c_3xe^{-2x}$$

Now, I want to continue from this point to solution of non-homogenous differential equation. Please give any hint or general solution!! Thanks in advance.

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up vote 4 down vote accepted

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 $$0+4\cdot 0+4A=2$$ $$\therefore A=\frac 12\implies y_p(x)=\frac x2$$

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

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HINT:

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


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

Find the complementary solution by solving:


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


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

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


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


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


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


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


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

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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 $w_p=y_p'$).

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1  
Perfect answer +1 – Yagna Patel Jan 14 at 20:44
    
Typo: you should have $w_p''$ at the start of the second line. – Ian Jan 14 at 21:01

Hint:

$$y'''+4y''+4y'=2$$ the particular solution should be (according to the undetermined Coefficient Method) $$y_p=A$$ but because there is a similirity with the complementary solution, so we should multiply by x $$y_p=Ax$$

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Your particular solution is Incorrect – Yagna Patel Jan 14 at 20:39
1  
@YagnaPatel, I have edited it, thanks for your comment – E.H.E Jan 14 at 21:02

we can make the D.E as a homogeneous D.E by taking the derivative $$y''''+4y'''+4y''=0$$ the characteristics equation $$r^2(r^2+4r+4)=0$$ so $$y=c_1+c_2x+c_3e^{-2x}+c_4xe^{-2x}$$

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