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jakeward123 March 9th, 2011 12:46 AM

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

aswoods March 9th, 2011 01:03 AM

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)...


jakeward123 March 9th, 2011 03:03 AM

Re: Exact form, differential equations
 
thanks for that!

for b) it is d/dx (ye^x)=e^(2x)

sorry about that.

aswoods March 9th, 2011 03:18 AM

Re: Exact form, differential equations
 
In that case you get , hence , a family of curves which includes cosh x (for C=).

jakeward123 March 9th, 2011 05:59 AM

Re: Exact form, differential equations
 
brilliant.

jakeward123 March 9th, 2011 07:56 AM

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..?

MarkFL March 9th, 2011 08:53 AM

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! :mrgreen:

jakeward123 March 10th, 2011 01:58 AM

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

MarkFL March 10th, 2011 06:49 AM

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.

jakeward123 March 10th, 2011 10:52 AM

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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