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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 March 14th, 2011, 12:19 PM #1 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Differential equations hey, wondering if this equation I should use substitution in the process or a different method. ie u=x/t find the general solution of the equation : (xt)dx/dt = x^2 + t^2 thanks March 14th, 2011, 12:40 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: Differential equations Yes, that is an appropriate substitution for this non-linear equation, leading to an implicit solution for x(t). March 15th, 2011, 10:41 AM #3 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Differential equations Can you check a these please 1. Find the general solutions of the first order equations. a) y dy/dx=x+2 ?y dy =?(x+2) dx y^2/2 = (1/2 x^2 +2x) +C y^2 = x^2 +4x + C y=sqrt (x^2 +4x + C) general solution b) dy/dx (y^2 -1) / x (1/y^2) -1 dy/dx = 1/x ?(1/y^2) -1 dy = ?1/x (1/y^2 -1) dy =?ln(x) +C y=tan(ln(x)+C) general solution c)x^2 dy/dx +2xy=2x d/dx(x^2 y)=2x exact form ?d/dx (x^2 y) dx = ?2x dx x^2 y = x^2 +C y=x^2 / x^2 + C/x^2 y=1+Cx^-2 general solution d) cosx dy/dx -y sinx = 1 d/dx (cosx y) =1 exact form ?d/dx (cosx y) dx =?1 dx cosx y = x y= x/cosx + C general solution March 15th, 2011, 11:17 AM #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: Differential equations a) You are correct up to the point: Next I would write it as: Complete the square on x: This is the family of "north-south" opening square hyperbolas centered at (-2,0) When you take the square root of both sides of an equation you need to put a � on one side, typically the side opposite the variable you are solving for. As you wrote the solution, you are only getting the top-half of the hyperbolas in the solution space. b) Separate variables: Use partial fraction decomposition on the left side: c) Correct. d) You need to put your constant of integration in place before you divide through by cos(x): March 15th, 2011, 12:18 PM #5 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Differential equations Thanks very much , although I was sure I had done well on those, never mind, your steps are good to follow.  March 15th, 2011, 12:40 PM #6 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Differential equations Ok, so i attempted my next question, hopefully i did better on this one. Find the particular solution of the first order differential equation x^2 dy/dx -y = 1 subject to y(2)=2 divide by x^2 to get equation in standard form dy/dx - 1/x y = 1/x^2 ?(x)=e^-?1/x dx =e^-?x^-2 = e^x^-1 multiply through by integrating factor e^x^-1 dy/dx - x^-2 y e^x^-1 = 1/x^2 e^x^-1 d/dx (e^x^1 y) = 1/x^2 e^x^-1 ?d/dx (e^x^-1 y) dx = ?1/x^2 e^x^-1 dx e^x^-1 y = 1/x^2 e^1/x e^1/x y = x^-2 e^1/x y= x^-2 e^1/x +C / e^1/x y=x^-2 = 1/x^2 + c/e^1/x general solution y(2)=2 2=1/2^2 + c/e^1/2 2=0.25+c/1.6487 (2-0.25) x 1.6487 =c 1.75 x 1.6487 = c 2.8358 =c y=1/x^2 + 2.8358 / e^1/x particular solution March 15th, 2011, 03:17 PM #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: Differential equations You have: Putting the equation in standard form, you should get: Calculate integrating factor: You have the correct integrating factor, so I assume your equation in standard form just contains a typo, where you have P(x) = 1/x. When you integrate the right side, you need to see that you initially have so you need to multiply the integral by a negative so that you get: In our case, Determine C from initial conditions: Thus the solution satisfying the given conditions is: March 15th, 2011, 05:18 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: Differential equations By the way, here is the working of the problem you gave in the first post: Divide through by xt: Use the substitution: =ut\:\therefore\:\frac {dx}{dt}=u+t\frac{du}{dt}" /> The equation is transformed to: March 16th, 2011, 03:26 AM #9 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Differential equations Great what about this one.. 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 = xe^x d/dx (x^-2 y) =xe^x exact form integrate wrt x d/dx(x^-2 y) dx = xe^x dx x^2 y = xe^x + e^x y= xe^x + e^x / x^2 y=e^x/x + e^x/x^2 general solution March 16th, 2011, 05:48 AM #10 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: Differential equations When you integrate: You get: Tags differential, equations ,
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