My Math Forum Solve the following ODEs...

 Calculus Calculus Math Forum

 February 14th, 2012, 08:02 AM #1 Senior Member   Joined: Feb 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, 10:18 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, 10:58 AM #3 Senior Member   Joined: Feb 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, 08:22 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 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, 08:55 AM #5 Senior Member   Joined: Feb 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, 01:29 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 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, 08:20 PM #7 Senior Member   Joined: Feb 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, 09:28 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 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, 03:22 PM #9 Global Moderator   Joined: Dec 2006 Posts: 21,119 Thanks: 2331 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, 04:52 PM   #10
Senior Member

Joined: Feb 2010

Posts: 324
Thanks: 0

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.

 Tags odes, solve

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post zibb3r Calculus 1 September 21st, 2013 02:35 PM Execross02 Applied Math 9 July 1st, 2013 09:15 AM singapore Calculus 2 March 19th, 2012 07:48 PM liakos Applied Math 0 January 1st, 2011 10:52 AM acnash Calculus 2 March 25th, 2010 05:19 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top