Answer: \(x=-38\mathrm{e}^{t/2}+6\mathrm{e}^{t/3}+t^2+11t+44\).
We solve
\(6\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}-5\dfrac{\mathrm{d}x}{\mathrm{d}t}+x=t^2+t+1\), with \(x(0)=12\) and \(x'(0)=-6\).
Complementary function
First solve the homogeneous equation
\(6x''-5x'+x=0\).
Its auxiliary equation is
\(6m^2-5m+1=0\).
Factorising,
\((3m-1)(2m-1)=0\), so the roots are \(m=\frac13\) and \(m=\frac12\).
Hence the complementary function is
\(x_c=A\mathrm{e}^{t/2}+B\mathrm{e}^{t/3}\).
Particular integral
Since the right-hand side is a quadratic, try
\(x_p=pt^2+qt+r\).
Then
\(x_p'=2pt+q\), and \(x_p''=2p\).
Substitute into the differential equation:
\(6(2p)-5(2pt+q)+(pt^2+qt+r)=t^2+t+1\).
Simplifying,
\(pt^2+(-10p+q)t+(12p-5q+r)=t^2+t+1\).
Equating coefficients gives
\(p=1\),
\(-10p+q=1\),
\(12p-5q+r=1\).
So
\(p=1\),
\(q=11\),
\(r=44\).
Therefore
\(x_p=t^2+11t+44\).
So the general solution is
\(x=A\mathrm{e}^{t/2}+B\mathrm{e}^{t/3}+t^2+11t+44\).
Use the initial conditions
Differentiate:
\(x'=\frac12A\mathrm{e}^{t/2}+\frac13B\mathrm{e}^{t/3}+2t+11\).
When \(t=0\),
\(x(0)=A+B+44=12\), so
\(A+B=-32\).
Also
\(x'(0)=\frac12A+\frac13B+11=-6\), so
\(\frac12A+\frac13B=-17\).
Multiply this second equation by 6:
\(3A+2B=-102\).
From \(A+B=-32\), we have \(2A+2B=-64\).
Subtracting,
\(A=-38\).
Then
\(B=6\).
Hence the particular solution satisfying the given conditions is
\(x=-38\mathrm{e}^{t/2}+6\mathrm{e}^{t/3}+t^2+11t+44\).
