Integrating differential equations
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From: Arno Narque on 19 Apr 2010 13:37 Hi guys! It's me again and again I am having a hard time with some %&$�/ differential equations again. I would be so grateful if somebody could give me a hint or oculd show me how to do this! Namely I should integrate the following equation: \dot{v}_{t}=pv_{t}+1-e^{p(\tau-t)} in the interval [0,\tau] where v_{0}=-\mu a_{0} and v_{\tau}=0 Could somebody please help me? Best regards, Arno
From: Arno Narque on 19 Apr 2010 13:40 On 2010-04-19 19:37:25 +0200, Arno Narque said: > Hi guys! > > It's me again and again I am having a hard time with some %&$�/ > differential equations again. I would be so grateful if somebody could > give me a hint or oculd show me how to do this! Namely I should > integrate the following equation: > > \dot{v}_{t}=pv_{t}+1-e^{p(\tau-t)} in the interval [0,\tau] > > where v_{0}=-\mu a_{0} and v_{\tau}=0 > > Could somebody please help me? > > Best regards, > > Arno A yeah and because the last time somebody told me I should tell people which solutions I've alrteady tried: I tried around with substitution but I have not really a clue how to apply this correctly since the t is in the power of e. merci! arno
From: Ray Vickson on 19 Apr 2010 15:19 On Apr 19, 10:37 am, Arno Narque <t16...(a)simonews.com> wrote: > Hi guys! > > It's me again and again I am having a hard time with some %&$§/ > differential equations again. I would be so grateful if somebody could > give me a hint or oculd show me how to do this! Namely I should > integrate the following equation: > > \dot{v}_{t}=pv_{t}+1-e^{p(\tau-t)} in the interval [0,\tau] By v_{t}, do you mean v(t)? Is dot{v}_{t} supposed to be v'(t) = dv(t)/ dt? I will assume so; then your DE reads as v'(t) = p*v(t) + 1 - c*exp(-p*t), where c = exp(p* tau). Actually, it is probably easier to change variables to v(t) = u(tau - t) = u(s), so your equation becomes u'(s) = -p*u(s) - 1 + exp(p*s), 0 <= s <= tau, with u(0) = 0 and u(tau) = -v0, where v0 is some constant (which happens to equal mu*a(0), but who cares, since we don't know what are mu and a(0)). An equation of the form u'(s) = -p*u(s) + f(s) (f(s) known) is entirely standard; it can be solved by using a so-called "integrating factor" or by consulting Chapter 1 of any DE textbook. R.G. Vickson > > where v_{0}=-\mu a_{0} and v_{\tau}=0 > > Could somebody please help me? > > Best regards, > > Arno
From: Arno Narque on 19 Apr 2010 18:26 On 2010-04-19 21:19:51 +0200, Ray Vickson said: > On Apr 19, 10:37�am, Arno Narque <t16...(a)simonews.com> wrote: >> Hi guys! >> >> It's me again and again I am having a hard time with some %&$�/ >> differential equations again. I would be so grateful if somebody could >> give me a hint or oculd show me how to do this! Namely I should >> integrate the following equation: >> >> \dot{v}_{t}=pv_{t}+1-e^{p(\tau-t)} in the interval [0,\tau] > > By v_{t}, do you mean v(t)? Is dot{v}_{t} supposed to be v'(t) = dv(t)/ > dt? I will assume so; then your DE reads as v'(t) = p*v(t) + 1 - > c*exp(-p*t), where c = exp(p* tau). Actually, it is probably easier to > change variables to v(t) = u(tau - t) = u(s), so your equation becomes > u'(s) = -p*u(s) - 1 + exp(p*s), 0 <= s <= tau, with u(0) = 0 and > u(tau) = -v0, where v0 is some constant (which happens to equal > mu*a(0), but who cares, since we don't know what are mu and a(0)). An > equation of the form u'(s) = -p*u(s) + f(s) (f(s) known) is entirely > standard; it can be solved by using a so-called "integrating factor" > or by consulting Chapter 1 of any DE textbook. > > R.G. Vickson > >> >> where v_{0}=-\mu a_{0} and v_{\tau}=0 >> >> Could somebody please help me? >> >> Best regards, >> >> Arno Wow thank you for this fast answer, you understood my notation completely right! I really surrendered this afternoon trying to solve this equation! Merci! Arno
From: James Burns on 19 Apr 2010 19:12
Arno Narque wrote: > On 2010-04-19 21:19:51 +0200, Ray Vickson said: > >> On Apr 19, 10:37 am, Arno Narque <t16...(a)simonews.com> wrote: >> >>> Hi guys! >>> >>> It's me again and again I am having a hard time with some %&$�/ >>> differential equations again. I would be so grateful if somebody could >>> give me a hint or oculd show me how to do this! Namely I should >>> integrate the following equation: >>> >>> \dot{v}_{t}=pv_{t}+1-e^{p(\tau-t)} in the interval [0,\tau] >> >> >> By v_{t}, do you mean v(t)? Is dot{v}_{t} supposed to be v'(t) = dv(t)/ >> dt? I will assume so; then your DE reads as v'(t) = p*v(t) + 1 - >> c*exp(-p*t), where c = exp(p* tau). Actually, it is probably easier to >> change variables to v(t) = u(tau - t) = u(s), so your equation becomes >> u'(s) = -p*u(s) - 1 + exp(p*s), 0 <= s <= tau, with u(0) = 0 and >> u(tau) = -v0, where v0 is some constant (which happens to equal >> mu*a(0), but who cares, since we don't know what are mu and a(0)). An >> equation of the form u'(s) = -p*u(s) + f(s) (f(s) known) is entirely >> standard; it can be solved by using a so-called "integrating factor" >> or by consulting Chapter 1 of any DE textbook. >> >> R.G. Vickson >> >>> >>> where v_{0}=-\mu a_{0} and v_{\tau}=0 >>> >>> Could somebody please help me? >>> >>> Best regards, >>> >>> Arno > > > Wow thank you for this fast answer, you understood my notation > completely right! I really surrendered this afternoon trying to solve > this equation! Something that may be causing you problems is that you have too many conditions on v(t). Either v_{0}=-\mu a_{0} or v_{\tau}=0 is enough to completely fix v(t), since the differential equation is first order. The other condition then places conditions on your constants, p, \tau, \mu, and a_{0} for there to be a solution. Jim Burns |