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 March 22nd, 2011, 04:33 PM #11 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: 2nd Order Differential Equation the (x^2) should be in there.. That comes from the y=c1 x^2 e^x +c2 e^-x. I see Ok great thanks. On another subject, if a problem d^2 x/dt^2 -4 dx/dt = 7-3e^4t d^2 x/dt^2 -4 dx/dt =0 , Auxillary equation, if it is of the form m^2 -4m =0, What would it be for m=? , as I realise if the auxillary equation is of the form m^2 -4 then, m= +or- 2. I should know this !
 March 22nd, 2011, 05:41 PM #12 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: 2nd Order Differential Equation We begin with: $\frac{d^2x}{dt^2}-4\frac{dx}{dt}=7-3e^{4t}$ Let's go another route with this one. Multiply through by $e^{-4t}$: $e^{-4t}\frac{d^2x}{dt^2}-4e^{-4t}\frac{dx}{dt}=7e^{-4t}-3$ Now, notice the left-side is the derivative of a product: $\frac{d}{dt}$$e^{-4t}\frac{dx}{dt}$$=7e^{-4t}-3$ $\int\,d$$e^{-4t}\frac{dx}{dt}$$=\int 7e^{-4t}-3\,dt$ $e^{-4t}\frac{dx}{dt}=-\frac{7}{4}e^{-4t}-3t+c_1$ Now, multiply through by $e^{4t}$ $\frac{dx}{dt}=-\frac{7}{4}-3te^{4t}+c_1e^{4t}$ $\int\,dx=\int -\frac{7}{4}-3te^{4t}+c_1e^{4t}\,dt$ Note: the 2nd term in the integrand on the right requires integration by parts. $x(t)=-\frac{7}{4}t-3$$\frac{1}{4}te^{4t}-\frac{1}{16}e^{4t}$$+\frac{1}{4}c_1e^{4t}+c_2$ $x(t)=e^{4t}$$\frac{3+4c_1}{16}-\frac{3t}{4}$$-\frac{7t}{4}+c_2$ Since $c_1$ is a constant, we may represent the constant $\frac{3+4c_1}{16}$ simply as $c_1$. $x(t)=e^{4t}$$c_1-\frac{3t}{4}$$-\frac{7t}{4}+c_2$ Now, had we worked this equation by the method of undetermined coefficients, we would first find: $x_h(t)=c_1+c_2e^{4t}$ Next, we would assume a particular solution of the form: $x_p(t)=Ate^{4t}+Bt$ (we need the t as a factor on both terms so that neither term is a form of $x_h(t)$) $x_p'(t)=A$$4te^{4t}+e^{4t}$$+B=Ae^{4t}$$4t+1$$ +B$ $x_p''(t)=A$$4e^{4t}+4(4t+1)e^{4t}$$=8Ae^{4 t}$$2t+1$$$ Now, substituting these into the original ODE gives: $$$8Ae^{4t}\(2t+1$$\)-4$$Ae^{4t}\(4t+1$$+B\)=7-3e^{4t}$ $16Ate^{4t}+8Ae^{4t}-16Ate^{4t}-4Ae^{4t}-4B=7-3e^{4t}$ $-4B+4Ae^{4t}=7-3e^{4t}$ Equating coefficients: $-4B=7\:\therefore\:B=-\frac{7}{4}$ $4A=-3\:\therefore\:A=-\frac{3}{4}$ Thus, we have: $x_p(t)=-\frac{3}{4}te^{4t}-\frac{7}{4}t$ So, putting it together, we have: $x(t)=x_h(t)+x_p(t)=$$c_1+c_2e^{4t}$$+$$-\frac{3}{4}te^{4t}-\frac{7}{4}t$$$ $x(t)=e^{4t}$$c_2-\frac{3t}{4}$$-\frac{7t}{4}+c_1$ Since the parameters are constants, we may write this in the form: $x(t)=e^{4t}$$c_1-\frac{3t}{4}$$-\frac{7t}{4}+c_2$
 March 24th, 2011, 07:08 AM #13 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: 2nd Order Differential Equation x dy/dx -2y = x^4 e^x Divide by x^3 x/x^3 dy/dx -2y/x^3 = x^4 e^x /x^3 x^-2 dy/dx -2x^-3 y = x e^x d/dx y/x^2 = x e^x. Exact form Integrate wrt x d(y/x^2) =x e^x dx y = x^2 (e^x (x-1) + c ) Can you point out my errors, I'm sure there will be some ! Thanks
 March 24th, 2011, 09:51 AM #14 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: 2nd Order Differential Equation No errors, looks good!
 March 26th, 2011, 02:01 AM #15 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: 2nd Order Differential Equation Ok so, say I had an equation similar to the 2nd order equation you helped me with before apart from the right side being slightly different. d^2 x/dt^2 -4 dx/dt = 7-3te^4t Very similar to previous one however there is '3t' in there now. My question is what form would the complimentry function xh(t) be and what could I try the particular solution xp(t) as. Could the right side be re-written or am I just looking to much into it? Jake
 March 26th, 2011, 04:58 AM #16 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: 2nd Order Differential Equation Since the left side is the same, the complimentary function would be the same: $x_h(t)=c_1+c_2e^{4t}$ For the particular solution, we could use: $x_p(t)=$$At^2+Bt$$e^{4t}+Ct$ Differentiating, substituting into the ODE, and solving for the undetermined coefficients gives: $x_p(t)=$$-\frac{3}{8}t^2+\frac{3}{16}t$$e^{4t}-\frac{7}{4}t$ Adding this to the complimentary function gives: $x(t)=\frac{1}{16}$$c_2+3t-6t^2$$e^{4t}-\frac{7}{4}t+c_1$
 March 26th, 2011, 07:14 AM #17 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: 2nd Order Differential Equation Awesome , but backing up bit I need to get the 1st and 2nd derivative right I'n my mind, I got x'p(t)= A(2+B)4e^4t + C and x''p(t)=8e^4t(A(B+2t)+2A(B+2t)) is this correct?
 March 26th, 2011, 11:44 AM #18 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: 2nd Order Differential Equation We have: $x_p(t)=$$At^2+Bt$$e^{4t}+Ct$ Using the product rule, we get (after simplification): $x_p'(t)=$$4At^2+(2A+4B)t+B$$e^{4t}+C$ $x_p''(t)=$$16At^2+(16A+16B)t+(2A+8B)$$e^{4 t}$
 March 26th, 2011, 12:47 PM #19 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: 2nd Order Differential Equation Ok I was way out. after subbing into the original equation, what is the end equation after simplfying.... Its a long one. So I can compare coefficients myself
 March 26th, 2011, 12:50 PM #20 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: 2nd Order Differential Equation After subbing into the original ODE and simplifying, I got: $$$8At+2A+4B$$e^{4t}-4C=7-3te^{4t}$

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