Tracker Posted January 23, 2012 Posted January 23, 2012 Solve the following differential equation by separation of variables: [math] \frac{xdy}{dx} = 4y [/math] My solution is: [math]\int \frac{dy}{4y} = \int \frac{dx}{x} \Rightarrow ln{\mid 4y \mid} = ln{\mid x \mid} + c [/math] The book's solution is: [math] y = cx^4 [/math] Can anyone show me how they came up this this solution? Thank you.
DrRocket Posted January 23, 2012 Posted January 23, 2012 Solve the following differential equation by separation of variables: [math] \frac{xdy}{dx} = 4y [/math] My solution is: [math]\int \frac{dy}{4y} = \int \frac{dx}{x} \Rightarrow ln{\mid 4y \mid} = ln{\mid x \mid} + c [/math] The book's solution is: [math] y = cx^4 [/math] Can anyone show me how they came up this this solution? Thank you. Take the log of both sides of the solution in the book.
Tracker Posted January 25, 2012 Author Posted January 25, 2012 Take the log of both sides of the solution in the book. [math] y = cx^4 \Rightarrow ln{y} = ln{cx^4} \rightarrow ln{y} = ln{c} + 4ln{x} \rightarrow ln{y} = 4ln{x} + c [/math] I don't understand how the four gets inside the ln to become [math] ln{4y} [/math] Thank you for the help. Cheers.
DrRocket Posted January 25, 2012 Posted January 25, 2012 [math]\int \frac{dy}{4y} = \int \frac{dx}{x} \Rightarrow \frac{1}{4} ln{\mid y \mid} = ln{\mid x \mid} + c [/math]
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