Solving Second order non - homogeneous Differential Equation

[ssh]

To Solve y’’ – 2 y’ – 3y = 64 e^-x x ---------------(1)
Using the method of undetermined coefficients :
The roots of the homogeneous equation are 3 and -1, so the complimentary solution is
y = c₁ e^3x + c₂ e^-x
Then the guess for the particular solution of (1) is e^-x x (Ax + B)
The Answer in the book is e^-x ( 8x² + 4x +1 ).
But there is no constant term in the guess now how does it show up in the answer?
Can anyone help me with this?
Thanks

 

[chisigma]

To Solve y’’ – 2 y’ – 3y = 64 e^-x x ---------------(1)
Using the method of undetermined coefficients :
The roots of the homogeneous equation are 3 and -1, so the complimentary solution is
y = c₁ e^3x + c₂ e^-x
Then the guess for the particular solution of (1) is e^-x x (Ax + B)
The Answer in the book is e^-x ( 8x² + 4x +1 ).
But there is no constant term in the guess now how does it show up in the answer?
Can anyone help me with this?
Thanks

The second order ODE is...

$\displaystyle y^{\ ''} -2\ y^{\ '} -3\ y = 64\ x\ e^{-x}$ (1)

... and you have to find a particular solution of the form...

$\displaystyle g(x)= (a\ x^{2} + b\ x +c)\ e^{-x}$ (2)

Deriving (2) You obtain...

$\displaystyle g^{\ '}(x)= \{ b - c + (2 a - b)\ x - a\ x^{2}\}\ e^{- x}$ (3)

$\displaystyle g^{\ ''}(x)= \{ 2 (a - b) + c + (b-4 a)\ x + a x^{2}\}\ e^{-x}$ (4)

Now insert in (1) the (2), (3) and (4) and then solve for a and b and c ...

Kind regards

$\chi$ $\sigma$

 

[Ackbach]

To Solve y’’ – 2 y’ – 3y = 64 e^-x x ---------------(1)
Using the method of undetermined coefficients :
The roots of the homogeneous equation are 3 and -1, so the complimentary solution is
y = c₁ e^3x + c₂ e^-x
Then the guess for the particular solution of (1) is e^-x x (Ax + B)
The Answer in the book is e^-x ( 8x² + 4x +1 ).
But there is no constant term in the guess now how does it show up in the answer?
Can anyone help me with this?
Thanks

The $e^{-x}\cdot 1$ part of your particular solution is unnecessary, since that exact piece shows up in the complementary solution. That is, you could just absorb that part into the complementary solution. It doesn't hurt anything, but it doesn't help. Your guess for the particular solution is the correct one. Moving along:

\begin{align*}
y_{p}&=e^{-x}(Ax^{2}+Bx)\\
y_{p}'&=-e^{-x}(Ax^{2}+Bx)+e^{-x}(2Ax+B)=e^{-x}(-Ax^{2}+(2A-B)x+B)\\
y_{p}''&=-e^{-x}(-Ax^{2}+(2A-B)x+B)+e^{-x}(-2Ax+2A-B)\\
&=e^{-x}(Ax^{2}+(-4A+B)x+2A-2B).
\end{align*}
Hence,
\begin{align*}
y_{p}''-2y_{p}'-3y_{p}&=e^{-x}(Ax^{2}+(-4A+B)x+2A-2B)\\
&\quad+e^{-x}(2Ax^{2}+(-4A+2B)x-2B)\\
&\quad+e^{-x}(-3Ax^{2}-3Bx)\\
&=e^{-x}(-8Ax+2A-4B).
\end{align*}
Set this equal to the RHS of the original DE, and I think you're good to go.

 

[ssh]

Thanks

 

[blue_raver22]

for your question. I can provide some insights into the solution of this problem. Here are some key points to consider:

1. The method of undetermined coefficients is a common approach for solving second order non-homogeneous differential equations. This method uses the roots of the homogeneous equation to find a particular solution.

2. In this particular case, the roots of the homogeneous equation are 3 and -1. This means that the complimentary solution is y = c1 e3x + c2 e-x.

3. To find the particular solution, we need to make a guess that includes the unknown coefficients A and B. In this case, the guess is e-x x (Ax + B).

4. The next step is to substitute this guess into the original equation and solve for the unknown coefficients A and B. This is where the constant term shows up in the answer. It is a result of solving for the unknowns A and B.

5. It is important to note that the constant term may not always appear in the guess, but it can still show up in the final solution. This is because it is a result of solving for the unknown coefficients.

In conclusion, the constant term in the solution is a result of solving for the unknown coefficients using the method of undetermined coefficients. I hope this explanation helps you understand the solution better. Let me know if you have any further questions.