My Math Forum  

Go Back   My Math Forum > College Math Forum > Differential Equations

Differential Equations Ordinary and Partial Differential Equations Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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
jakeward123 is offline  
 
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 ex and not e^(2x)...

aswoods is offline  
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.
jakeward123 is offline  
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=).
aswoods is offline  
March 9th, 2011, 05:59 AM   #5
Senior Member
 
Joined: Feb 2011

Posts: 150
Thanks: 0

Re: Exact form, differential equations

brilliant.
jakeward123 is offline  
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..?
jakeward123 is offline  
March 9th, 2011, 08:53 AM   #7
Senior Member
 
MarkFL's Avatar
 
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!
MarkFL is offline  
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
jakeward123 is offline  
March 10th, 2011, 06:49 AM   #9
Senior Member
 
MarkFL's Avatar
 
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.
MarkFL is offline  
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
jakeward123 is offline  
Reply

  My Math Forum > College Math Forum > Differential Equations

Tags
differential, equations, exact, form



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
First-Order, Exact and Linear Differential Equations summsies Differential Equations 2 July 24th, 2013 05:24 PM
2 exact differential equation helloprajna Differential Equations 12 July 19th, 2013 09:21 PM
Exact differential equation Survivornic Differential Equations 3 October 1st, 2012 07:02 PM
Non-Exact Differential equation Survivornic Differential Equations 2 September 30th, 2012 02:40 PM
Solutions to exact equations conjecture Calculus 0 July 27th, 2008 10:25 AM





Copyright © 2019 My Math Forum. All rights reserved.