Question

Cosx log(2y-8)+1/x]dx+(sinx/y-4)dy=0

Answer 1

We are to solve a first-order nonlinear ordinary differential equation:

$$\bigg(\text{cos}x\cdot\text{log}(2y-8)+\frac{1}{x}\bigg)\text{d}x+\frac{\text{sin}x}{y-4}\text{d}y=0.$$

Open up the parentheses:

$$\text{cos}x\cdot\text{log}(2y-8)\text{d}x+\frac{\text{d}x}{x}+\frac{\text{sin}x}{y-4}\text{d}y=0.\space\space\space\space\space\space(1)$$

After thinking for a while we can see that

$$\frac{\text{d}x}{x}=\text{d}(\text {log}x),\\ \space\\ \text{cos}x\text{d}x=\text{d}(\text {sin}x),$$

$$\frac{\text{d}y}{y-4} = \text{d}(\text{log}(2y-8)).$$

Therefore, if we make a substitution in equation (1):

\text{cos}x=f(x),\space\frac{1}{x}=g(x),\\ \space\\ [f(x)g(x)]'=f'(x)g(x)+g'(x)f(x)

(chain rule),

we can rewrite equation (1) above as

$$d(\text{sin}x)\cdot\text{log}(2y-8)+\text{d(log}x)+\text{sin}x\cdot\text{d}(\text{log}(2y-8))=0,\\$$

finally, integration by parts gives

$$y(x)=\frac{1}{2}x^{-\frac{1}{\text{sin}x}}\big(e^{\frac{c_1}{\text{sin}x}}+8x^{\frac{1}{\text{sin}x}}\big).$$