My Math Forum Exact form, differential equations

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 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) $\int \mathrm{d}(yx^2)= \int x^3 \mathrm{d}x\\yx^2 = \frac14x^4+A\\y=\frac14x^2+\frac{A}{x^2}$ b) I take it that it's definitely e²x and not e^(2x)... $\int \mathrm{d}(ye^x)= e^2 \int x \mathrm{d}x\\ye^x=\frac12e^2x^2+C\\y=\frac{\frac12 e^2x^2+C}{e^x}$
 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 $ye^x= \frac12e^{2x}+C$, hence $y= \frac12e^x+Ce^{-x}$, 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: $e^{2x}\frac{dy}{dx}+2e^{2x}y=x^2$ The left side is the differentiation of the product $e^{2x}y$ with respect to x: $\frac{d}{dx}$$e^{2x}y$$=x^2$ $\int\,d$$e^{2x}y$$=\int x^2\,dx$ $e^{2x}y=\frac{1}{3}x^3+C$ $y=e^{-2x}$$\frac{1}{3}x^3+C$$$ b) $x\frac{dy}{dx}+y=x^3$ $\frac{d}{dx}$$xy$$=x^3$ $\int\,d(xy)=\int x^3\,dx$ $xy=\frac{1}{4}x^4+C$ $y=\frac{1}{4}x^3+Cx^{-1}$ That's how I would solve those, however, you are instructed to express the equations in exact form: a) $e^{2x}\frac{dy}{dx}+2e^{2x}y=x^2$ $$$2e^{2x}y-x^2$$dx+$$e^{2x}$$dy=0$ Test for exactness: $\frac{\delta}{\delta y}$$2e^{2x}y-x^2$$=2e^{2x}$ $\frac{\delta}{\delta x}$$e^{2x}$$=2e^{2x}$ Since $\frac{\delta}{\delta y}$$2e^{2x}y-x^2$$=\frac{\delta}{\delta x}$$e^{2x}$$$ the equation is exact. Thus, we have $\frac{\delta F}{\delta x}=2e^{2x}y-x^2$ and integrating with respect to x, we have: $F(x,y)=e^{2x}y-\frac{1}{3}x^3+g(y)$ Now we take the partial derivative with respect to y: $\frac{\delta F}{\delta y}=e^{2x}+g#39;(y)$ Since $\frac{\delta F}{\delta y}=e^{2x}$ we have $g'(y)=0$ thus $g(y)=C$ giving: $F(x,y)=e^{2x}y-\frac{1}{3}x^3+C$ The solution is given implicitly by: $F(x,y)=C$ thus we have: $e^{2x}y=\frac{1}{3}x^3+C$ $y=e^{-2x}$$\frac{1}{3}x^3+C$$$ 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 (y-x^3) dx + x dy =0 Test for exactness d/dy(y-x^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: $\frac{\delta}{\delta y}$$y-x^3$$=1$ $\frac{\delta}{\delta x}(x)=1$ 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

 Tags differential, equations, exact, form

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post summsies Differential Equations 2 July 24th, 2013 05:24 PM helloprajna Differential Equations 12 July 19th, 2013 09:21 PM Survivornic Differential Equations 3 October 1st, 2012 07:02 PM Survivornic Differential Equations 2 September 30th, 2012 02:40 PM conjecture Calculus 0 July 27th, 2008 10:25 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top