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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 nonlinear 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 "northsouth" 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 tophalf 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 (20.25) 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: 

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