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# Answer to Question #101860 in Electrical Engineering for Sahil Patel

Question #101860
Cosx log(2y-8)+1/x]dx+(sinx/y-4)dy=0
1
2020-01-28T07:10:18-0500

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),\\\\\n\\space\\\\\n\\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),\\\\\n\\space\\\\\n[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)."

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