My Math Forum Solve the following ODEs...

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 February 14th, 2012, 07:02 AM #1 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Solve the following ODEs... 1. (5y-(5/2)xy^3) dx -dy =0 2. (-3x+y+6)dx + (x+y+dy) = 0 Yea I got these totally wrong.. can I get some pointers?
 February 14th, 2012, 09:18 AM #2 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408 Re: Solve Hello, FreaKariDunk! There is a typo in #2. $\text{Is this correct?}\;(-3x\,+\,y\,+\,6)\,\!dx \,+\,(x\,+\,y)\,\!dy \:=\:0\,?$ $\text{Assuming that it is, we have: }\:(x\,+\,y)\,\!dy \;=\;(3x\,-\,y\,-\,6)\,\!dx$ The equation is Reducible to Homogenous . . . a rather tedious procedure. $\text{Let: }\:\begin{Bmatrix}u &=& x\,+\,y && \Rightarrow && du &=& dx\,+\,dy \\ \\ \\ v &=& 3x\,-\,y\,-\,6 && \Rightarrow && dv &=& 3dx\,-\,dy \end{Bmatrix}$ $\text{Solve for }dx\text{ and }dy:\;\begin{Bmatrix}dx=&\frac{du\,+\,dv}{4} \\ \\ \\ dy=&\frac{3du\,-\,dv}{4} \end{Bmatrix}=$ $\text{Substitute: }\:u\left(\frac{3du\,-\,dv}{4}\right) \;=\; v\left(\frac{du\,+\,dv}{4}\right) \;\;\;\Rightarrow\;\;\; 3u\,\!du\,-\,u\,\!dv \:=\:v\,\!du\,+\,v\,\!dv$ [color=beige]. . . . . . . . . [/color]$u\,\!dv\,+\,v\,\!dv \:=\:3u\,\!du\,-\,v\,\!du \;\;\;\Rightarrow\;\;\; (u\,+\,v)\,\!dv \:=\:(3u\,-\,v)\,du$ [color=beige]. . . . . . . . . [/color]$\frac{dv}{du} \:=\:\frac{3u\,-\,v}{u\,+\,v} \;\;\;\Rightarrow\;\;\;\frac{dv}{du}\:=\:\frac{3-\frac{v}{u}}{1\,+\,\frac{v}{u}}\;\;\;\text{ Homogeneous!}$ $\text{Let} w \,=\,\frac{v}{u} \;\;\;\Rightarrow\;\;\;v \:=\:wu \;\;\;\Rightarrow\;\;\;\frac{dv}{du} \:=\:w\,+\,u\,\!\frac{dw}{du}$ $\text{Substitute: }\:w\,+\,u\,\!\frac{dw}{du} \:=\:\frac{3\,-\,w}{1\,+\,w} \;\;\;\Rightarrow\;\;\;u\,\!\frac{dw}{du} \:=\:\frac{3\,-\,w}{1\,+\,w}\,-\,w$ $\text{And we have: }\:u\,\!\frac{dw}{du} \:=\:\frac{3\,-\,2w\,-\,w^2}{1\,+\,w}$ $\text{Separate the variables! }\;\frac{1\,+\,w}{3\,-\,2w\,-\,w^2}\,\!dw \:=\:\frac{du}{u}$ $\text{Can you finish it now?}$
 February 14th, 2012, 09:58 AM #3 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Re: Solve It is supposed to be (-3x + y + 6)dx - (x+y+2)dy = 0 Sorry.
 February 14th, 2012, 07:22 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Solve 1.) $$$5y-\frac{5}{2}xy^3$$dx-dy=0$ We can express this in the form: $\frac{dy}{dx}+(-5)y=$$-\frac{5}{2}x$$y^3$ and we now have a Bernoulli equation with n = 3, $P(x)=-5$ and $Q(x)=-\frac{5}{2}x$. To transform this ODE into a linear equation, we first divide by $y^3$ to obtain: $y^{-3}\frac{dy}{dx}-5y^{-2}=-\frac{5}{2}x$ Next, we make the substitution $v=y^{-2}$. Since $\frac{dv}{dx}=-2y^{-3}\frac{dy}{dx}$, the transformed equation is: $-\frac{1}{2}\frac{dv}{dx}-5v=-\frac{5}{2}x$ or $\frac{dv}{dx}+10v=5x$ Now we have a linear equation, for which the integrating factor is $\mu(x)=e^{10x}$: $e^{10x}\frac{dv}{dx}+10e^{10x}v=5xe^{10x}$ $\frac{d}{dx}$$e^{10x}v$$=5xe^{10x}$ Integrating with respect to x yields: $\int\,d$$e^{10x}v$$=5\int xe^{10x}\,dx$ $e^{10x}v=\frac{1}{20}e^{10x}$$10x-1$$+c_1$ $v=\frac{10x-1}{20}+c_1e^{-10x}$ Back substituting for v, we have the implicit relation: $y^{-2}=\frac{10x-1}{20}+c_1e^{-10x}$ Not included in the last equation is the solution $y\equiv0$ that was lost in the process of dividing by $y^3$ above. 2.) $$$-3x+y+6$$dx+$$x+y+2$$dy=0$ I am assuming you mean this based on your initial post. This is an equation with linear coefficients and we will use the translation of axes $x=u+h,\,y=v+k$ where h and k satisfy the system: $-3h+k+6=0$ $h+k+2=0$ Solving this system, we find h = 1 and k = -3. Hence, we let: $x=u+1$ $y=v-3$. Since $dy=dv$ and $dx=du$ substitution into the ODE for x and y yields $$$-3u+v$$du+$$u+v$$dv=0$ or $\frac{dv}{du}=\frac{3-\frac{v}{u}}{1+\frac{v}{u}}$ The above equation is homogeneous, so we let $z=\frac{v}{u}$, and so: $\frac{dv}{du}=z+u\cdot\frac{dz}{du}$ and substituting for $\frac{v}{u}$ we obtain: $z+u\frac{dz}{du}=\frac{3-z}{1+z}$ Separating variables gives: $\int\frac{z+1}{z^2+2z-3}\,dz=-\int\frac{1}{u}\,du$ $\frac{1}{2}\ln\|z^2+2x-3\|=-\ln|u|+c_1$ from which it follows that: $z^2+2z-3=Cu^{-2}$ When we substitute back in for z, u, and v, we find: $$$\frac{v}{u}$$^2+2$$\frac{v}{u}$$-3=Cu^{-2}$ $v^2+2uv-3u^2=C$ $(v+3u)(v-u)=C$ $$$(y+3)+3(x-1)$$$$(y+3)-(x-1)$$=C$ $(3x+y)(y-x+4)=C$ This last equation gives an implicit solution to the original ODE.
 February 15th, 2012, 07:55 AM #5 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Re: Solve the following ODEs... Thanks for your help Mark. Btw, you're always picking on me. :P I couldn't find my thread because you changed the name. Thanks for clarifying! You're my favorite mod... along with the others.
 February 15th, 2012, 12:29 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Solve the following ODEs... Awww...I didn't mean to pick on you!! I simply felt that this topic deserved a more descriptive title than "Solve."
 February 15th, 2012, 07:20 PM #7 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Re: Solve the following ODEs... Solve worked because I needed a quick subject so I could have valentines lunch with my White Trash Wife. It's hard trying to be as awesome as you.
 February 15th, 2012, 08:28 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Solve the following ODEs... What was awesome was finding both of these problems as worked out examples in my old DiffEq textbook!
 February 17th, 2012, 02:22 PM #9 Global Moderator   Joined: Dec 2006 Posts: 20,820 Thanks: 2159 The guessed equation can simply be integrated to give -(3/2)x² + 6x + xy + (1/2)y² = c. The corrected equation can simply be integrated to give -(3/2)x² + 6x + xy + (1/2)y² + 2y = c. The LHS of the second of the above solutions can be factorized, giving (3x + y)(-x + y + 4)/2 = c. When c = 0, this gives (3x + y)(-x + y + 4)/2 = 0, but this should be converted to the solutions y = -3x and y = x - 4.
February 17th, 2012, 03:52 PM   #10
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Re: Solve the following ODEs...

Quote:
 Originally Posted by MarkFL What was awesome was finding both of these problems as worked out examples in my old DiffEq textbook!

So uh... lemme borrow that pleaseee? I would love to find the solutions manual to the text book he gave us online.

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