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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

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)...

$\int \mathrm{d}(ye^x)= e^2 \int x \mathrm{d}x\\ye^x=\frac12e^2x^2+C\\y=\frac{\frac12 e^2x^2+C}{e^x}$

 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)

 aswoods March 9th, 2011 03:18 AM

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).

 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:

$e^{2x}\frac{dy}{dx}+2e^{2x}y=x^2$

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! :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:

$\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.

 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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