Answer: \(y=\left(1+\frac{7}{x}\right)\ln\left(\frac{x+7}{8}\right)+\frac{7}{x}\).

First put the differential equation into linear form by dividing through by \(x(x+7)\):

\(\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{7}{x(x+7)}y=\frac{1}{x+7}.\)

Use partial fractions:

\(\frac{7}{x(x+7)}=\frac{1}{x}-\frac{1}{x+7}.\)

So the integrating factor is

\(\mu(x)=\exp\left(\int \frac{7}{x(x+7)}\,\mathrm{d}x\right)=\exp\left(\int \left(\frac{1}{x}-\frac{1}{x+7}\right)\,\mathrm{d}x\right)=e^{\ln x-\ln(x+7)}=\frac{x}{x+7}.\)

Multiply the equation through by \(\frac{x}{x+7}\):

\(\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{xy}{x+7}\right)=\frac{x}{(x+7)^2}.\)

Rewrite the right-hand side as

\(\frac{x}{(x+7)^2}=\frac{1}{x+7}-\frac{7}{(x+7)^2}.\)

Integrating gives

\(\frac{xy}{x+7}=\int \left(\frac{1}{x+7}-\frac{7}{(x+7)^2}\right)\,\mathrm{d}x=\ln(x+7)+\frac{7}{x+7}+C.\)

Use \(y=7\) when \(x=1\):

\(\frac{1\cdot 7}{1+7}=\ln 8+\frac{7}{8}+C,\)

so \(\frac{7}{8}=\ln 8+\frac{7}{8}+C\), hence \(C=-\ln 8\).

Therefore

\(\frac{xy}{x+7}=\ln(x+7)-\ln 8+\frac{7}{x+7}=\ln\left(\frac{x+7}{8}\right)+\frac{7}{x+7}.\)

Now multiply through by \(\frac{x+7}{x}\):

\(y=\frac{x+7}{x}\ln\left(\frac{x+7}{8}\right)+\frac{7}{x}=\left(1+\frac{7}{x}\right)\ln\left(\frac{x+7}{8}\right)+\frac{7}{x}.\)