Agohofidov6

2022-01-21

Particular solution to $y{}^{″}-3{y}^{\prime }+2y=2{e}^{x}$

limacarp4

Expert

$y{}^{″}-3{y}^{\prime }+2y=2{e}^{x}$.
Use operator D:
Let $D=\frac{d}{dx}$.
So, $\left({D}^{2}-3D+2\right){y}_{p}=2{e}^{x}⇒\left(D-1\right)\left(D-2\right){y}_{p}=2{e}^{x}⇒$
${y}_{p}=\frac{1}{\left(D-1\right)\left(D-2\right)}\cdot 2{e}^{x}$
$=2{e}^{x}\frac{1}{\left(D+1-1\right)\left(D+1-2\right)}\cdot 1=2{e}^{x}\frac{1}{D\left(D-1\right)}\cdot 1⇒$
${y}_{p}=2{e}^{x}\frac{1}{D\left(0-1\right)\cdot 1}=-2{e}^{x}\frac{1}{D}\cdot 1=-2x{e}^{x}$

ambarakaq8

Expert

This is too long for a comment, so I posted it as an answer. First solve for the homogeneous equation $y{}^{″}-3{y}^{\prime }+2y=0$ by setting the right hand side to be zero. The auxiliary equation is ${m}^{2}-3m+2=0$, which has roots $m=2,1$. Therefore the solution for this homogeneous equation is . Now we want to find a particular solution $y{}^{″}-3{y}^{\prime }+2y=2{e}^{x}$. Normally we set the particular solution to be $A{e}^{x}$. However, it duplicates with the solution of the homogeneous solution, therefore, we multiple it with x until no duplication occurs. Therefore, the particular solution is given by $Ax{e}^{x}$.
Let me do another example: to solve $y{}^{″}-2{y}^{\prime }+y=2{e}^{x}$.
First solve the homogenous equation $y{}^{″}-2{y}^{\prime }+y=0$. The auxiliary equation is ${m}^{2}-2m+1=0$ which has double roots $m=1$. Therefore, the solution for this homogeneous equation is . Now if we want to find a particular solution $y{}^{″}-2{y}^{\prime }+y=2{e}^{x}$. Normally we set the particular solution to be $A{e}^{x}$. However, it duplicates with the solution of the homogeneous solution, therefore, we multiple it with x and it becomes $Ax{e}^{x}$, but it still duplicates with $x{e}^{x}$. Therefore, we mupltiply it by ${x}^{2}$, and the particular solution is given by $A{x}^{2}{e}^{x}$.

RizerMix

Expert

The simplest thing in your case is probably to use this (easily justified) trick: If p(x) is a one variable polynomial with complex coefficients and c a complex number, then the ODE $p\left(\frac{d}{dx}\right)y={e}^{cx}$ has a solution of the form ${e}^{cx}q\left(x\right)$ where q(x) is a polynomial whose degree is the number of roots of p(x) equal to c. In you case, $p\left(x\right)=\left(x-1\right)\left(x-2\right),c=1$, so the trick tells you to look for a solution of the form $\left(ax+b\right){e}^{x}$, and you can clearly assume $b=0$.