Find a family of solutions to the DE dy/dx+ 2xy=1

I have some trouble separating the variables...anyone can give me a general method on how to do that?

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- Jan 16th 2010, 06:25 AMdaskywalkerSimple differential equation
Find a family of solutions to the DE dy/dx+ 2xy=1

I have some trouble separating the variables...anyone can give me a general method on how to do that? - Jan 16th 2010, 06:41 AMCalculus26
The trouble is this is not a separable equation.

You have to use an integrating factor .

Do you know this method ? - Jan 16th 2010, 07:13 AMdaskywalker
oh sorry let me rephrase the question:

Verify that the indicated family of functions is a solution to the DE dy/dx+2xy=1:

y= (e^-x^2)*integral from x to 0 (e^(t^2)dt)+ce^(-x^2)

where c is an arbitrary constant.

I suppose the e^t^2 is the integrating factor??

I tried to substitute the y equation into the DE but it does not work. - Jan 16th 2010, 07:40 AMCalculus26
See the attachment

By the way The integrating factor would be e^(x^2) but you'll get to that later I assume - Jan 16th 2010, 08:22 AMchisigma
If You write the DE as...

$\displaystyle y^{'} = -2xy +1$ (1)

... it is easy to see that it's a*linear*DE on the form...

$\displaystyle y^{'} = a(x)\cdot y +b(x)$ (2)

... where $\displaystyle a(x)= -2x$ and $\displaystyle b(x)=1$ and its solution can be found is 'standard fashion' ...

$\displaystyle y= e^{\int a(x)\cdot dx} \{\int b(x)\cdot e^{-\int a(x)\cdot dx }\cdot dx +c\} = e^{-x^{2}}\cdot (\int e^{x^{2}}\cdot dx + c)$ (3)

The integral in (3) is not elementary but it can be expressed as a convergent series...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$