Tracker Posted January 23, 2012 Share 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. Link to comment Share on other sites More sharing options...
DrRocket Posted January 23, 2012 Share 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. Link to comment Share on other sites More sharing options...
Tracker Posted January 25, 2012 Author Share 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. Link to comment Share on other sites More sharing options...
DrRocket Posted January 25, 2012 Share 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] Link to comment Share on other sites More sharing options...
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