# Consider d y </mrow> d x </mrow> </mfrac>

Consider
$\frac{dy}{dx}+p\left(x\right)y={\int }_{0}^{\mathrm{\infty }}y\left(x\right)dx.$
I want to solve above differential equation. Can I consider right hand side as constant to solve this?
I know RHS is a constant but it also involves solution y, which might create trouble unless solution is known to us.
Also is it possible to find solution to given ordinary differential equation which is independent of y.
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Jerome Page
You are correct in considering the right side first as a mostly independent constant. Then you can apply the integration formula
${y}^{\prime }+py=C\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}y\left(x\right)=y\left(0\right){e}^{-P\left(x\right)}+C{\int }_{0}^{x}{e}^{P\left(s\right)-P\left(x\right)}ds.$
Now with this you can return to the original equation to try to determine C
$C={\int }_{0}^{\mathrm{\infty }}y\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx=y\left(0\right){\int }_{0}^{\mathrm{\infty }}{e}^{-P\left(x\right)}\phantom{\rule{thinmathspace}{0ex}}dx+C{\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{x}{e}^{P\left(s\right)-P\left(x\right)}ds\phantom{\rule{thinmathspace}{0ex}}dx$
Now if the integrals involved have finite values, and if then C does not cancel from the equation, you can get a value for C that in addition is proportional to y(0) (as the linearity of the original equation demands).