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solving Second order non - homogeneous Differential Equation

ssh

New member
Jun 30, 2012
17
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 e3x + 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

Well-known member
Feb 13, 2012
1,704
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 e3x + 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

Indicium Physicus
Staff member
Jan 26, 2012
4,193
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 e3x + 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.
 
Last edited:

ssh

New member
Jun 30, 2012
17
Thanks