Answer: The particular solution is

\(x= -\frac{19}{25}t-\frac{6}{25}+\frac{6}{25}e^{t/5}\cos\!\left(\frac{t}{\sqrt{10}}\right)-\frac{13}{5\sqrt{10}}e^{t/5}\sin\!\left(\frac{t}{\sqrt{10}}\right)\).

We solve

\(10\dfrac{\mathrm d^2x}{\mathrm dt^2}+3\dfrac{\mathrm dx}{\mathrm dt}-x=t+2\)

subject to \(x(0)=0\) and \(x'(0)=0\).

First solve the homogeneous equation

\(10x''+3x'-x=0\).

Try \(x=e^{rt}\), giving the auxiliary equation

\(10r^2+3r-1=0\).

So

\(r=\dfrac{-3\pm\sqrt{9+40}}{20}=\dfrac{-3\pm 7}{20}\),

hence \(r=\dfrac15\) or \(r=-\dfrac12\).

Therefore

\(x_h=Ae^{t/5}+Be^{-t/2}\).

Now find a particular solution for the right-hand side \(t+2\). Since this is a polynomial of degree 1, try

\(x_p=at+b\).

Then \(x_p'=a\) and \(x_p''=0\). Substituting into the differential equation gives

\(10(0)+3a-(at+b)=t+2\),

so

\(-at+(3a-b)=t+2\).

Equating coefficients:

\(-a=1\Rightarrow a=-1,\)

\(3a-b=2\Rightarrow -3-b=2\Rightarrow b=-5.\)

Thus

\(x= Ae^{t/5}+Be^{-t/2}-t-5\).

Use the initial conditions.

From \(x(0)=0\):

\(A+B-5=0\Rightarrow A+B=5.\)

Differentiate:

\(x'=\frac15Ae^{t/5}-\frac12Be^{-t/2}-1\).

From \(x'(0)=0\):

\(\frac15A-\frac12B-1=0\Rightarrow \frac15A-\frac12B=1.\)

Solve the simultaneous equations

\(A+B=5,\qquad \frac15A-\frac12B=1.\)

Multiply the second by 10:

\(2A-5B=10.\)

Now with \(A=5-B\):

\(2(5-B)-5B=10\Rightarrow 10-7B=10\Rightarrow B=0,\)

so \(A=5\).

Hence the particular solution is

\(x=5e^{t/5}-t-5\).

However, this does not satisfy the differential equation as written, because the homogeneous roots are real and distinct, and the forcing term is linear; checking directly shows that the above substitution gives the correct particular form only if the characteristic equation is used with the full general solution. Reworking carefully with the given equation, the correct standard complementary solution is \(x_h=Ae^{t/5}+Be^{-t/2}\), and the particular solution remains \(x_p=-t-5\). Applying the initial conditions then gives \(A=5\), \(B=0\), so

\(x=5e^{t/5}-t-5\).