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March 9th, 2011, 12:46 AM  #1 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Exact form, differential equations
hey, Ive had a go at this question but im not too sure. Solve the following differential equations given in exact form Qa) d/dx (yx^2)=x^3 Qb)d/dx (ye^x)=e^2x Aa) integrate wrt x. ?d/dx (yx^2)dx=?x^3 dx yx^2=x^4/4+A divide by x^2 to get y> y=x^2/4 +A Ab) integrate wrt x. ?d/dx(ye^x)dx=?e^2x ye^x=e^2 x^2 /2 divide by e^x to get y> y=( 1/2(e^2 x^2) )/ e^x Jake 
March 9th, 2011, 01:03 AM  #2 
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: Exact form, differential equations
Look after the constants, as they help you find general solutions. a) b) I take it that it's definitely e²x and not e^(2x)... 
March 9th, 2011, 03:03 AM  #3 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Exact form, differential equations
thanks for that! for b) it is d/dx (ye^x)=e^(2x) sorry about that. 
March 9th, 2011, 03:18 AM  #4 
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: Exact form, differential equations
In that case you get , hence , a family of curves which includes cosh x (for C=½).

March 9th, 2011, 05:59 AM  #5 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Exact form, differential equations
brilliant.

March 9th, 2011, 07:56 AM  #6 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Exact form, differential equations
Ok so, By expressing the following differential equations in exact form, find their general solutions. a) e^2x dt/dx + 2e^2x y = x^2 b) x dy/dx + y = x^3 The first one when writing in standard form, dividing by the exponential, do they cancel out leaving e^2x dx as the integrating factor? and for the second one I end up with e^ln(x) .... = x being the integrating factor..? 
March 9th, 2011, 08:53 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: Exact form, differential equations
I am assuming a) is: The left side is the differentiation of the product with respect to x: b) That's how I would solve those, however, you are instructed to express the equations in exact form: a) Test for exactness: Since the equation is exact. Thus, we have and integrating with respect to x, we have: Now we take the partial derivative with respect to y: Since we have thus giving: The solution is given implicitly by: thus we have: I'll leave this method for b) to you! 
March 10th, 2011, 01:58 AM  #8 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Exact form, differential equations
Thanks Mark, testing for exactness the process just confirms that the equation is exact, producing the same general solution, when testing for exactness on b) xdy/dx +y=x^3 (yx^3) dx + x dy =0 Test for exactness d/dy(yx^3) = 1/2 x^2 d/dx(x) = 1/2 x^2 Is this correct so far 
March 10th, 2011, 06:49 AM  #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: Exact form, differential equations
No, you would have: When taking the partial derivative of a function of more than one variable, you treat the other other variables (those you are not differentiating with respect to) as a constant. 
March 10th, 2011, 10:52 AM  #10 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Exact form, differential equations
ok then, so I followed your method but I could only get y  1/4x^4 +C => y=1/4x^4+Cx^1 , but it does not match the solution as previously stated y=1/4 x^3 +Cx^1


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