Exercises 2.2⇒x(1+Cet)=t⇒x=tt.Note: WhenCis replaced by−K, this answer can also be written asx=Ket/(t−1).Further we observe that since we divide byx−x2=x(1−x), thenx≡0andx≡1arealsosolutions. AllowingKto be zero givesx≡0, but no choice forKwill givex≡1, so we listthis as a separate solution.15.To separate variables, we move the term containindxto the right-hand side of the equationand divide both sides of the result byy. This yieldsy−1dy=−yecosxsinxdx⇒y−2dy=−ecosxsinxdx.Integrating the last equation, we obtainZy−2dy=Z(−ecosxsinx)dx⇒−y−1+C=Zeudu(u=cosx)1y+C=eu=ecosx⇒y=1C−ecosx,whereCis an arbitrary constant.17.First we ±nd a general solution to the equation. Separating variables and integrating, we getdydx=x3(1−
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.