Question #101860

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

Expert's answer

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

Open up the parentheses:

After thinking for a while we can see that

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

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

(chain rule),

we can rewrite equation (1) above as

finally, integration by parts gives

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