Variation of Parameters

**VonNemo19** · March 1st 2011, 10:02 AM

Hello. I'm having a bit of trouble understanding the rationale behind the assumption that is made in the variation of parameters method for first and second order linear differential equations.

For the first order equations, my book lays out the property of linear equations that the solution $y$ is the sum of the complementary and a particular solution, that is $y=y_c+y_p$ . Then the author shows that the homogeneous equation

$\frac{dy}{dx}+P(x)y=0$

is seperable and has as its solution $y_c=ce^{-\int{P(x)}dx}$ and letting $e^{-\int{P(x)}dx}=y_1(x)$ for convenience, we have $y_c=cy_1(x)$ .

Now the author states that in order to find the particular solution $y_p$ we use a procedure known as variation of parameters where we assume that $y_p=uy_1$ .

So, I have no trouble at all with the derivation from here on. My question is, how is it that this assumption is made. Or, stated in another way, what was the reasoning behind making this assumption. What is it that would make one believe that the particular solution would be simply some function of the independent variable times the complementary solution?

**Random Variable** · March 1st 2011, 10:21 AM

I think it's just because we don't want to look for particular solution that has a complicated form. The proof shows that when you apply certain conditions, the form of the particular solution is valid in most (if not all) cases.

**TheEmptySet** · March 1st 2011, 10:22 AM

Originally Posted by VonNemo19

Hello. I'm having a bit of trouble understanding the rationale behind the assumption that is made in the variation of parameters method for first and second order linear differential equations.

For the first order equations, my book lays out the property of linear equations that the solution $y$ is the sum of the complementary and a particular solution, that is $y=y_c+y_p$ . Then the author shows that the homogeneous equation

$\frac{dy}{dx}+P(x)y=0$

is seperable and has as its solution $y_c=ce^{-\int{P(x)}dx}$ and letting $e^{-\int{P(x)}dx}=y_1(x)$ for convenience, we have $y_c=cy_1(x)$ .

Now the author states that in order to find the particular solution $y_p$ we use a procedure known as variation of parameters where we assume that $y_p=uy_1$ .

So, I have no trouble at all with the derivation from here on. My question is, how is it that this assumption is made. Or, stated in another way, what was the reasoning behind making this assumption. What is it that would make one believe that the particular solution would be simply some function of the independent variable times the complementary solution?

In my opinion the reasoning it that if you make this assumption of the product form of the solution, after differentiating (via the product rule) you will always end up with a factor of
$u(x)$ multiplied by the homogeneous ODE which is always equal to 0. The hope then is that this new ODE for $u$ would be solvable.

So as in your example suppose $y_c$ solve the homogeneous system
$y'+p(x)y=0$ and you want to solve

$y'+p(x)y=f(x)$

Using this assumption as you get
$y=uy_c \implies y'=u'y_c+uy_c'$ now putting this into the ode gives

$\displaystyle u'y_c+uy_c'+p(x)uy_c=f(x) \iff u'y_c+\underbrace{u[y_c'+p(x)y_c]}_{\text{homogeneous ODE solution}=0}=f(x) \iff u'y_c=f(x) \iff u'=\frac{f(x)}{y_c}$

**topsquark** · March 1st 2011, 10:29 AM

Or it could simply be the old Physics derivation stand-by... It works.

-Dan

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