MHB 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