Question

Use the integrating factor to solve the differential equation. dz/dy = z tan y + sin y

Answer 1

$\cfrac{dz}{dy}=z \cdot tan(y)+sin(y) \\ \cfrac{dz}{dy}-z \cdot tan(y)=sin(y)$

Then, when we compare the form to $\cfrac{dz}{dy}+P \cdot z=Q$, we can identify that the integrating factor will be: $I=e^{\int Pdy}=e^{-\int tan(y)dy}=e^{\ln(cos(y))}=cos(y)$

Then, we multiply the integrating factor I and proceed to solve the equation:

$\\ (cos(y))[\cfrac{dz}{dy}-z \cdot tan(y)]=sin(y)cos(y) \\ cos(y)\cfrac{dz}{dy}-z \cdot sin(y)=sin(y)cos(y) \\ \cfrac{d}{dy}(z \cdot cos(y))=sin(y)cos(y) \iff d(z \cdot cos(y))= sin(y)cos(y){dy} \\ \int d(z \cdot cos(y))=\int sin(y)cos(y){dy} \\ z \cdot cos(y)=-\frac{1}{2} cos^2(y)+C$

We divide the last result by cos(y) to determine z:

$z =C\cdot sec(y)-\cfrac{1}{2} cos(y)$

In conclusion, we determine that $z =C\cdot sec(y)-\cfrac{cos(y)}{2}$ .