Homogeneous differential equation, show integral =1

Question

I can't understand and therefore can't get started nor solve the question in the photo. It is from the Cambridge IB DP Math HL Calculus Option book and should take about 15 minutes to solve. click here for a photo of the question

Answer 1

$$\int_k^2\frac {dy}{f(y)}=1$$ Note that

$$1=\ln(e)-\ln(1)=\int_1^e\frac {dx}{x}=\int_{\nu(1)}^{\nu(e)}\frac {d\nu}{f(\nu)}=\int_k^2\frac {d\nu}{f(\nu)}$$ $$1=\int_k^2\frac {d\nu}{f(\nu)}=\int_k^2\frac {dy}{f(y)}$$ Since we have that: $$\nu(e)=2 \text{ and } \nu(1)=k$$ For question b

Use a substitution $y=tx$ then $y'=t'x+t$ To evaluate the integral

$$y'=\ln(y)-\ln(x)=\ln(tx)-\ln(x)=\ln(t)+\ln(x)-\ln(x)=\ln(t) $$ $$\implies t'x=\ln(t)-t \implies \int \frac {dt}{\ln(t)-t}=\int \frac {dx}x$$