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I'm trying an easy problem to get my bearings using the method here.

The integral is $\int_3^5{\frac{x^2}{1+x^2}dx}$.

I would like to proceed, if possible to solve by defining:

$F(y) = \int_3^5{\frac{\sin{(y\cdot x})}{1+x^2}dx}$

Then obtaining $-F''(y) = \int_3^5{\frac{x^2\sin{(y\cdot x})}{1+x^2}dx}$

Adding gives $F(y) - F''(y)= \int_3^5{\frac{(1+x^2)\sin{(y\cdot x})}{1+x^2}dx}$

or $F(y) - F''(y) - \frac{\cos{3y}-\cos{5y}}{y} = 0.$

This is where I'm stuck. I don't know how to solve the differential equation. Any help would be greatly appreciated. I'm assuming that this can be done.

  • 0
    That should be $u-u''$ in my above comment. I'd expand in an answer but not feeling up to it right now. Cheers,2011-09-06

1 Answers 1

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Let's observe the next differential equation:

y''-y+ \frac{\cos3x-\cos5x}{x}=0 ,

This is nonhomogeneous second-order ordinary differential equation of the form: y''+p(x)y'+q(x)y-g(x)=0 ,which can be solved if the general solution to the homogenous version is known, in which case variation of parameters can be used to find the particular solution:

$y_p=-y_1(x)\int\frac{y_2(x)g(x)}{W(x)}\, dx + y_2(x)\int\frac{y_1(x)g(x)}{W(x)}\, dx$ , where $y_p$ is particular solution, $y_1(x)$ and $y_2(x)$ are the homogeneous solutions of the equation y''+p(x)y'+q(x)y=0 and W(x) is the Wronskian of these two functions.