Answer: \(\frac{dy}{dx}=\frac12e^{2x}-\frac1{x+1}+\frac52\); \(y=\frac14e^{2x}-\ln(x+1)+\frac52x+\frac{15}{4}\).

(a) Integrate

\(\frac{d^2y}{dx^2}=e^{2x}+\frac1{(x+1)^2}.\)

This gives

\(\frac{dy}{dx}=\frac12e^{2x}-(x+1)^{-1}+C.\)

Since \(\frac{dy}{dx}=2\) when \(x=0\),

\(2=\frac12e^0-1+C.\)

So

\(2=\frac12-1+C,\)

which gives

\(C=\frac52.\)

Therefore

\(\frac{dy}{dx}=\frac12e^{2x}-\frac1{x+1}+\frac52.\)

(b) Integrate again:

\(y=\frac14e^{2x}-\ln(x+1)+\frac52x+D.\)

Since \(y=4\) when \(x=0\),

\(4=\frac14-\ln1+0+D.\)

Thus

\(D=\frac{15}{4}.\)

Hence

\(y=\frac14e^{2x}-\ln(x+1)+\frac52x+\frac{15}{4}.\)