Question

Find the particular solution of ydx - 4dy = x ^ 2 dy when x = 4 and y = 1

Answer 1

$$ydx-4dy=x^2dy$$

$$(x^2+4)dy=ydx$$

$$\dfrac{dy}{y}=\dfrac{dx}{x^2+4}$$

Integrate

$$\int\dfrac{dy}{y}=\int\dfrac{dx}{x^2+4}$$

$$\ln|y|=\dfrac{1}{2}\tan^{-1}(x/2)+C$$

When $x = 4$ and $y = 1$

$$\ln|1|=\dfrac{1}{2}\tan^{-1}(4/2)+C$$

$$C=-\dfrac{1}{2}\tan^{-1}(2)$$

The particular solution is

$$\ln|y|=\dfrac{1}{2}\tan^{-1}(x/2)-\dfrac{1}{2}\tan^{-1}(2)$$