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Given the differential equation $L \ u = f$, with $$L \ u = a_2(t) \frac{d^2u}{dt^2}+a_1(t)\frac{du}{dt}+a_0(t)u$$ with $a_i(t)$ sufficiently smooth and $a_2(t) \neq 0$ for every $t$. Suppose that $u_0$ is a solution to the homogenenous equation $L \ u=0$, with $u_0(t)\neq0$ for every $t$. How can I then use the variation of parameters method by Lagrange to find the solution $u =u_0v$ ? And how can I find the general solution if the equation isn't homogeneous?

Update: I've substituted $u=u_0v$ into the equation and found a new $2^{nd}$ order differential equation of the form $$f = \frac{d^2v}{dt^2}(a_2(t)u_0)+\frac{dv}{dt}(2a_2(t)u_0'+a_1(t)u_0)+va_0(t)u_0$$ which I'm supposed to solve for $w$ using $w=v'$. But I don't seem to be able to proceed as I'm not sure how to solve the system $$\begin{equation} \begin{split} w =&\ v'\\ w'a_2(t)u_0+w(2a_2(t)u_0'+a_1(t)u_0)+v(a_0(t)u_0)=&\ f\ \\ \end{split} \end{equation}$$

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2 Answers 2

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Here is how we can solve the homogeneous equation $ L u = 0$. Once we have both solutions of this equation, we can use the method of variation of parameters to find a solution to $ L u = f $.

First we calculate derivatives of the solution $ u = u_0 v $ that you suggested: $$ u' = u_0' v + u_0 v' $$ $$ u'' = u_0'' v + 2 u_0' v' + u_0 v'' $$ Next, because $ u = u_0 v $ is a solution to the homogeneous equation $ a_2(t) u'' + a_1(t) u' + a_0(t) u = 0 $ we have: $$ a_2(t) (u_0'' v + 2 u_0' v' + u_0 v'') + a_1(t) (u_0' v + u_0 v') + a_0(t) u_0 v = 0 $$ We can re-arrange the terms and the equation in terms of $ v $ and its derivatives: $$ a_2(t) u_0 v'' + (2 a_2(t) u_0' + a_1(t) u_0) v' + (a_2(t) u_0'' + a_1(t) u_0' + a_0(t) u_0) v = 0 $$ Now, because $ u_0 $ satisfies the equation $ L u = 0 $ , the coefficient of $ v $ is zero, and the equation becomes: $$ a_2(t) u_0 v'' + (2 a_2(t) u_0' + a_1(t) u_0) v' = 0 $$ This looks like a second order equation, but it can be solved as a first order equation, if we take $ w = v' $ : $$ a_2(t) u_0 w' + (2 a_2(t) u_0' + a_1(t) u_0) w = 0 $$ From here, we solve this equation for $ w $, calculate the integral of $ w $ to find $ v $, and multiply $ v $ by $ u_0 $ to find the solution $ u $.

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With

$$ L(u) = u''+\frac{a_1}{a_2} u'+\frac{a_0}{a_2} u=u''+b_1 u'+b_0 u $$

given a non null $u_0(t)$ such that $L(u_0) = 0$ being a second order $L$, we can obtain another independent solution $L(u_1) = 0$ as follows: if $c_0$ is a generic constant then $L(c_0u_0) = 0$ so choose $u_1 = c_0(t)u_0$ then we have

$$ L(c_0(t)u_0) = c_0(t)L(u_0) + u_0c''_0+(b_1u_0+2u'_0)c'_0=u_0c''_0+(b_1u_0+2u'_0)c'_0=0 $$

making $\phi = c'_0$ we have a first order DE as $\frac{\phi'}{\phi} + \frac{u_0}{(b_1u_0+2u'_0)}=0$ so obtaining $\phi\to u_1$ is direct.

Having $u_0, u_1$ for the homogeneous case, let us approach the case $L(u) = \frac{f}{a_2(t)}$. The method of variation of constants now will be applied to

$$ L(c_0(t)u_0+c_1(t)u_1)=\frac{f}{a_2(t)} $$

so developing

$$ L(c_0(t)u_0+c_1(t)u_1)=c_0L(u_0)+c_1L(u_1) + u_0c''_0+(b_1u_0+2u'_0)c'_0+c''_1u_1+(b_1u_1+2u'_1)c'_1-\frac{f}{a_2(t)}=0 $$

Here as $c_0, c_1$ are independent we can choose them as

$$ \cases{ u_0c''_0+(b_1u_0+2u'_0)c'_0-\frac{f}{a_2(t)}=0\\ c''_1u_1+(b_1u_1+2u'_1)c'_1=0 } $$

finally determining $c_1(t), c_2(t)$ for the non-homogeneous case.

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