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Problem 4 Medium Difficulty
Solve the differential equation.
$ y' + xe^y = 0 $
Answer
$y=-\ln \left(\frac{x^{2}}{2}+C\right)$
Topics
Differential Equations
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Top Calculus 2 / BC Educators
Catherine R.
Missouri State University
Kayleah T.
Harvey Mudd College
Kristen K.
University of Michigan - Ann Arbor
Samuel H.
University of Nottingham
Watch More Solved Questions in Chapter 9
Video Transcript
this question us is to solve the differential equation. Why one plus X each the power of why is equivalent to zero first things first. We want to write this in terms of D y DX. We know why one is the same thing as D y detox. This is set equal to negative X each the why it's negative because we'll bring it over to the right hand side. Now we know we want to write this just with the wise on one side and the exes on one side. So do a little bit of manipulation to write. Eat the negative Y Do y is equivalent to negative acts. D Backs again acts on one side, wise on one side. Take the integral of both sides and we have e to the negative fly is X squared over two plus c. Don't forget the constant integration. Now take the natural walk of both sides. We know that Ln times e is just one. We have negative. Why is natural log of X squared over two plus c last step. We want this in terms of positive. Why not negative? Why
Topics
Differential Equations
Top Calculus 2 / BC Educators
Catherine R.
Missouri State University
Kayleah T.
Harvey Mudd College
Kristen K.
University of Michigan - Ann Arbor
Samuel H.
University of Nottingham