A method for finding a particular solution of a non-homogeneous differential equation is given by the variation of parameters formula.

This resembles the reduction of order method in the sense that we make an assumption on the form the solution $x=x_{1}v$ and then solve for $v$.

How do we really know that this is a solution in the end? We only said,

If $x$ satisfy the equation and is of the form $x=v_{1}x_{1}+v_{2}x_{2}$ then $v_{1}=..$ and $v_{2}=..$ making it nessesary but not sufficent.

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    $\begingroup$ Not sure exacltly what you mean: The rule that holds is that the solution is given by the sum of the homogeneous solution and the particular solution. Here you are adding a restriction to the possibile solution space of the inhomogeneous equation hoping that this will indeed result in a solvable and easier equation. It might also be that doing the variation of parameters formula you end up with no solution at all, but this doesn't mean that the equation has no solution. $\endgroup$ – b00n heT Aug 14 '18 at 11:52
  • $\begingroup$ @b00n heT right, and the only way of knowing wether it is a solution is plugging it in? My text seem to indicate that this gives a solution no matter what, which I feel is a little odd. $\endgroup$ – user561840 Aug 14 '18 at 13:46
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    $\begingroup$ @b00nheT Actually, variation of parameters (if done right) always guarantees finding a particular solution of non-homogeneous linear differential equation (or system of such equations). The trick is that you end up with a system $W(t) v'(t) = \alpha(t)$, where $v = (v_1, \dots, v_n)$ is unknown vector of "constants", $W(t)$ is a Wronskian matrix for some fundamental system of solutions that you choose and $\alpha(t)$ is some known right hand side. Since Wronskian is always non-zero for a fundamental system of solutions, it is possible to solve this system for each $t$. $\endgroup$ – Evgeny Aug 14 '18 at 14:10
  • $\begingroup$ @Evengy ok, but during the course of doing this we also impose the restriction that our solution satisfy the equation. How is this then a procedure that leads to a solution every time? In a way we assume that $x=vy$ is a solution and conclude that $v$ must be of a certain form. $\endgroup$ – user561840 Aug 14 '18 at 14:20
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    $\begingroup$ Yes. This ansatz always works and the general solution of non-homogeneous equation can always be written in the form that is used in method of parameter variation. $\endgroup$ – Evgeny Aug 16 '18 at 7:13

The idea is simple, if we have an equation of the form $$xy+y'=g(x) \quad \text{ for exemple}$$ In general, it is not easy to find a closed form of the solution directly, so we try to solve a similar equation but easier which is the homogeneous one $$xy+y'=0$$ Hence one has a solution $y_h(x)=ce^{-\frac{x^2}{2}}$ for this last equation, now we think about the non-homogeneous equation which her solution seems to be pretty close to $y_h$. So let's variate $y_h$ to be its solution, now we can try to find a solution of the form $y_g(x)=v(x)y_h(x)$ or $y_g(x)=y_h(x) +v(x)$ the choice is up to you (up to the simplicity of the calculs), here we choice $y_g(x)=y_h(x) +v(x)$ so

$$ xy_g(x)+y_g'(x)=cxe^{-\frac{x^2}{2}} +xv(x)-cxe^{-\frac{x^2}{2}}+v'(x)=g(x) $$ $$ \Rightarrow xv(x) +v'(x)=g(x)$$ $$ \Rightarrow (xv(x) +v'(x))e^{\frac{x^2}{2}}=g(x)e^{\frac{x^2}{2}}$$ $$ \Rightarrow \frac{d}{dx}(v(x)e^{\frac{x^2}{2}})=g(x)e^{\frac{x^2}{2}}$$

which can be solvable mostly, then we get the correction factor which we use to transform $y_h$ into a solution of the non-homogenous equation!

I hope that this can help


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