The ODE
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\cos^2\left(\dfrac{\pi y}4\right)[/tex]
is separable, as
[tex]\dfrac{\mathrm dy}{\cos^2\left(\frac{\pi y}4\right)}=\mathrm dx[/tex]
[tex]\sec^2\left(\dfrac{\pi y}4\right)\,\mathrm dy=\mathrm dx[/tex]
Integrate both sides:
[tex]\displastyle\int\sec^2\left(\dfrac{\pi y}4\right)\,\mathrm dy=\int\mathrm dx[/tex]
In the left integral, substitute u = πy/4 and du = π/4 dy :
[tex]\displastyle\frac4\pi\int\sec^2(u)\,\mathrm du=\int\mathrm dx[/tex]
[tex]\dfrac4\pi \tan(u)=x+C[/tex]
[tex]\dfrac4\pi\tan\left(\dfrac{\pi y}4\right)=x+C[/tex]
Given that y = 1 when x = 0, we have
[tex]\dfrac4\pi\tan\left(\dfrac\pi4\right)=C\implies C=\dfrac4\pi[/tex]
since tan(π/4) = 1. So the ODE has a particular solution of
[tex]\dfrac4\pi\tan\left(\dfrac{\pi y}4\right)=x+\dfrac4\pi[/tex]
or
[tex]\tan\left(\dfrac{\pi y}4\right)=\dfrac{\pi x}4+1[/tex]
Now when y = 3, we have
[tex]\tan\left(\dfrac{3\pi}4\right)=\dfrac{\pi x}4+1[/tex]
[tex]-1=\dfrac{\pi x}4+1[/tex]
[tex]-2=\dfrac{\pi x}4[/tex]
[tex]-8=\pi x[/tex]
[tex]x=-\dfrac8\pi[/tex]
making the answer C.
