Question

Find the particular solution of `r (dr)/(dθ) + cos θ` = 5 at r = `sqrt2` and θ = 0.

Answer 1

`r (dr)/(dθ) + cos θ` = 5

∴ `r (dr)/(dθ) = 5 - cos θ`

∴ r dr = (5 − cos θ) dθ

Integrating both sides, we get

`int r  dr = int (5-cosθ) dθ`

∴ `r^2/2 = 5θ - sinθ + c`

This is the general solution.

Now, θ = 0 and r = `sqrt2` we get

∴ `(sqrt2)^2/2 = 5(0) - sin0 + c`

∴ c = 1

∴ The particular solution is `r^2/2 = 5θ - sinθ + 1`.