Answer: The particular solution is \(y=-(49+56x)e^{-x/7}+49(x+1)\).

The auxiliary equation is

\(49m^2+14m+1=0\),

so

\(\left(7m+1\right)^2=0\),

and therefore \(m=-\frac{1}{7}\) is a repeated root. Hence the complementary function is

\(y_c=(A+Bx)e^{-x/7}.\)

For a particular integral, try \(y_p=px+q\). Then \(y_p'=p\) and \(y_p''=0\). Substituting in

\(49y''+14y'+y=49x+735\)

gives

\(14p+px+q=49x+735.\)

Comparing coefficients gives \(p=49\) and \(14p+q=735\), so \(q=49\). Thus

\(y=(A+Bx)e^{-x/7}+49x+49.\)

Using \(y(0)=0\),

\(A+49=0\), so \(A=-49\).

Now

\(y'=\left(B-\frac{A+Bx}{7}\right)e^{-x/7}+49.\)

Using \(y'(0)=0\),

\(B-\frac{A}{7}+49=0.\)

With \(A=-49\), this gives \(B+7+49=0\), so \(B=-56\).

Therefore

\(\boxed{y=-(49+56x)e^{-x/7}+49(x+1)}.\)