I am not 100% sure what is "steady-state solution w(x)". I tried to solve the question by first making equation (1) equals to zero; then integrate it twice with respect to x and substituted u=0,x=0 and u=H, x=L into the equation and ended up with A=GL/(2c^2) where A is an arbitrary constant. So do my steps make sense? Am I answering what the question is asking me? Many Thanks.
The problem posed is the first (and easy) part of a tougher problem. You start with a string which is flat at $t=0$ and is not moving at that time. Then instataneously one side of the string is raised to $H$, producing a delta-function pulse in the string, and thereafter the ends are held at $0$ and $H$ and the tension is such that the wave-speed is $c$. As $t$ grows, that pulse position will start to move toward $x=0$ and the motion can be computed by Fourier decomposing the pulse in terms of the string's harmonics, and superposing those wave motions. The problem might then ask you to show that as $t$ goes to infinity the solution approaches the steady state solution you found in step (a).
That steady state solution is easy enough to find: Since $u(x,t)$ does not depend on $t$, you have $$ u(L) = H\\u(0) = 0\\ \frac{d^2 u}{dx^2} = -\frac{G}{c^2} $$ The latter gives a general solution $$ u = -\frac{Gx^2}{2c^2} + ax + b$$ and since $u(0)=0$ we know that $b=0$. Then $$ H = -\frac{GL^2}{2c^2}+aL \Longrightarrow a=\frac{H}{L}+\frac{GL}{2c^2}$$ The desired solution is $$ u = -\frac{Gx^2}{2c^2}+\left( \frac{H}{L}+\frac{GL}{2c^2}\right)x $$