Hello f1player:
The method of undetermined coefficients.
\(\displaystyle \L\\y"-2y'-3y=x^{2}-1\)
The roots of \(\displaystyle \L\\m^{2}-2m-3=0\) are 3 and -1.
So we have the complementary solution:
\(\displaystyle \L\\y_{c}=C_{1}e^{3x}+C_{2}e^{-x}\)
Because the right hand side is a quadratic we can use the form
\(\displaystyle \L\\Ax^{2}+Bx+C\)
\(\displaystyle \L\\y_{p}=Ax^{2}+Bx+C\)
\(\displaystyle \L\\y'_{p}=2Ax+B\)
\(\displaystyle \L\\y''_{p}=2A\)
Sub into the equation:
\(\displaystyle \L\\2A-2(2Ax+B)-3(Ax^{2}+Bx+C)=x^{2}-1\)
\(\displaystyle \L\\-3Ax^{2}-4Ax-3Bx+2A-2B-3C=x^{2}-1\)
Equating coefficients gives the system:
\(\displaystyle \L\\-3A=1\)
\(\displaystyle \L\\-4A-3B=0\)
\(\displaystyle \L\\2A-2B-3C=-1\)
Solving the system gives:
\(\displaystyle \L\\A=\frac{-1}{3}, \;\ B=\frac{4}{9}, \;\ C=\frac{-5}{27}\)
Which gives the particular solution:
\(\displaystyle \L\\y_{p}=\frac{-1}{3}x^{2}+\frac{4}{9}x-\frac{5}{27}\)
Therefore, the general solution is \(\displaystyle y=y_{c}+y_{p}\)
\(\displaystyle \L\\\fbox{y=C_{1}e^{3x}+C_{2}e^{-x}-\frac{1}{3}x^{2}+\frac{4}{9}x-\frac{5}{27}}\)
See?.