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 May 29th, 2012, 03:53 PM #1 Member   Joined: May 2012 Posts: 34 Thanks: 0 help me, differential equation When mixing a certain alcoholic beverage, a barperson starts with a 10 litre solution of 20% alcohol. Into this solution the barperson mixes a 40% alcohol mixture at the rate of 1 litre/minute and pours into glasses at the rate of 2 litres/minute. (a) Find a formula for the volume V (t) of the mixture at time t minutes. (b) Find a differential equation that tells the rate of change of the amount of alcohol A(t) at time t minutes. (c) Solve the differential equation and use the initial condition to find a formula for A(t) dont know hard qs, can u help me
 May 29th, 2012, 09:59 PM #2 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: help me, differential equation There's no need to bump a topic after a few hours. It will be seen and someone will come along and help. If after a few days no one does, then its okay to bump it, preferably with some work or thoughts on how to proceed.
 May 29th, 2012, 10:02 PM #3 Member   Joined: May 2012 Posts: 34 Thanks: 0 Re: help me, differential equation sorry but i am doing my practice now and want to finish
 May 29th, 2012, 10:04 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: help me, differential equation Post your thoughts about what you should do, and when I log on tomorrow, I promise I will help.
 May 29th, 2012, 10:42 PM #5 Member   Joined: May 2012 Posts: 34 Thanks: 0 Re: help me, differential equation ok i wait for u
 May 30th, 2012, 04:43 AM #6 Member   Joined: May 2012 Posts: 34 Thanks: 0 Re: help me, differential equation This is as good as i can think please help me dy/dt= the rate of flow in - rate of flowing out so flow in is 2/5 litre alcohol says the question and flow out is 2y(t)? so dy/dt+2y(t)=2/5?
 May 30th, 2012, 07:18 AM #7 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: help me, differential equation See this topic. Apparently someone is studying from the same book you are.
 May 30th, 2012, 11:37 AM #8 Member   Joined: May 2012 Posts: 34 Thanks: 0 Re: help me, differential equation i tried but my work is very big confusion can u show me steps to do part a.b.c
 May 30th, 2012, 12:13 PM #9 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: help me, differential equation a) Let V be measured in liters and t be measured in minutes. We are given the IVP: $\frac{dV}{dt}=1-2=-1$ where $V(0)=10$ Hence, we have: $\int\,dV=-\int\,dt$ $V(t)=-t+C$ We may now use the given initial condition to determine C: $V(0)=C=10$ thus: $V(t)=10-t$ b) Let A be measured also in liters. We are given: $\frac{dA}{dt}=\frac{2}{5}-2\cdot\frac{A(t)}{V(t)}$ where $A(0)=2$ Write the ODE in standard linear form: $\frac{dA}{dt}+\frac{2}{10-t}A=\frac{2}{5}$ c) Now we compute the integrating factor: $\mu(x)=e^{\int\frac{2}{10-t}\,dt}=e^{\ln$$\frac{1}{(t-10)^2}$$}=\frac{1}{(10-t)^2}$ Multiplying the ODE by the integrating factor gives: $\frac{1}{(10-t)^2}\frac{dA}{dt}+\frac{2}{(10-t)^3}A=\frac{2}{5}\cdot\frac{1}{(10-t)^2}$ $\frac{d}{dt}$$\frac{A}{(10-t)^2}$$=\frac{2}{5(10-t)^2}$ $\int\,d$$\frac{A}{(10-t)^2}$$=\frac{2}{5}\int\frac{1}{(10-t)^2}\,dt$ $\frac{A}{(10-t)^2}=\frac{2}{5(10-t)}+C$ $A(t)=\frac{2}{5}(10-t)+C(10-t)^2$ We may now use the given initial condition to determine C: $4+100C=2$ $C=-\frac{1}{50}$ thus: $A(t)=\frac{2}{5}(10-t)-\frac{1}{50}(10-t)^2=\frac{10-t}{50}$$20-(10-t)$$=\frac{10-t}{50}$$10+t$$=\frac{100-t^2}{50}=2-\frac{1}{2}$$\frac{t}{5}$$^2$

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