Question

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

Answer 1

$$\dfrac{dz}{dy}-z\tan y=\sin y$$

The integrating factor: $\mu(y)=\cos y$

$$\cos y\dfrac{dz}{dy}-z(\cos y)(\tan y)=\cos y\sin y$$

$$\dfrac{d}{dy}((\cos y)z)=-z\sin y+\cos y\dfrac{dz}{dy}$$

Then

$$\dfrac{d}{dy}((\cos y)z)+z\sin y-z\sin y=\cos y\sin y$$

$$d((\cos y)z)=\cos y\sin ydy$$

Integrate

$$\int d((\cos y)z)=\int\cos y\sin ydy$$

$$(\cos y)z=\dfrac{1}{2}\sin ^2y+C$$

$$z=\dfrac{\sin ^2y}{2\cos y}+\dfrac{C}{\cos y}$$