2
$\begingroup$

Consider the non-homogeneous ODE

$$y'' + p(x)y' + q(x)y = g(x), g(x)\neq0$$

I do not understand how come the homogeneous equation $Y_h(x)$ and particular solution $Y_p(x)$ to the non-homogeneous equation can be linearly independent.

$\endgroup$
1
$\begingroup$

Suppose $Y_h$ is a nonzero solution of the homogeneous equation and $Y_p$ is a solution of the non homogeneous equation. Suppose also that $\alpha Y_h+\beta Y_p=0$. Differentiating twice gives \begin{gather} \alpha Y_h+\beta Y_p=0\\ \alpha Y_h'+\beta Y_p'=0\\ \alpha Y_h''+\beta Y_p''=0 \end{gather} Multiply the first equation by $q$, the second equation by $p$ and sum, getting $$ \alpha(Y_h''+pY_h'+qY_h)+\beta(Y_p''+pY_p'+qY_p)=0 $$ and therefore $$ \beta g=0 $$ that gives $\beta=0$. Since $Y_h\ne0$, we also have $\alpha=0$.

Note that the hypothesis $Y_h\ne0$ is essential.

$\endgroup$
  • $\begingroup$ Sorry, I don't get the part where βg=0. Can u explain to me how you yield that step? Thank you. $\endgroup$ – Duck Mar 28 '16 at 1:39
  • $\begingroup$ @Duck If $g$ is a nonzero function, you have $g(t)\ne0$ for some $t$. Now $\beta g(t)=0$, so... $\endgroup$ – egreg Mar 28 '16 at 8:01
  • $\begingroup$ Oh right... okay I got it. Thank you very much! $\endgroup$ – Duck Mar 28 '16 at 11:45
0
$\begingroup$

the space of the solution is the whole space of continuous functions. now consider the space of homogeneous solution witch is a subspace of space of continuous functions. if the particular solution is not in this space then simply they are independent which means if u cant write the particular solution in terms of homogeneous solution. in this case overall solution is simply solution to

if $g(x)$ can be written in terms of $Y_h(x)$ then you must find a solution witch can't be spanned by $Y_h(x)$ but still satisfy the equation or else if $Y_p(x)$ has no difference with $Y_h(X)$ and left side of equation is always equal to zero.

when $p, q\; and \;g$ are continuous this particular solution always exist in other words there is a space orthogonal to $Y_h(x)$ which contains $Y_p(x)$ but finding $Y_p(x)$ in close form is not always possible. but finding a particular solution with other approximate ways like power series expansion with Frobenius Method is possible.

for understanding this concept easiest way is to consider the simple case of Linear ODE with constant coefficients: $$y'' + 3y' + 2y = sin(x)$$ in this ODE $Y_h(x)=c_1e^{-x}+c_2e^{-2x}$ and $Y_p(x)=c_3sin(x)+c_4cos(x)$ and you see that particular solution can not be spanned by homogeneous solution space basis witch are $\{ e^{-x},e^{-2x} \}$

now consider this ODE: $$y'' + 3y' + 2y = e^{-x}$$ in this case still $Y_h(x)=c_1e^{-x}+c_2e^{-2x}$ but for particular solution we can't consider $y_p(x)=c_3e^{-x}$ cause this function is homogeneous subspace spanned by $\{ e^{-x},e^{-2x} \}$. so we have to find another subspace of the space of continuous functions which satisfy the equation and yet is orthogonal to space of $y_h(x)$. hopefully in this case it's easy to verify that the space spanned by the function $y_p(x)=cxe^{-x}$ is suitable.

when speaking of ODE's with constant coefficients it's easy to find $y_h(x)$ and $y_p(x)$ .Also you can get Another nice interpretation with the aid of Laplace Transform and the notion of Pole's and Zero's. but for example if we consider the Bessel's equation: $$x^2y''+xy'+(x^2-v^2)=g(x)$$ finding the solution spaces in closed form is not possible anymore.

Also it's worth mentioning that the difference of two particular solution's Always lie's in the homogeneous subspace in other words:

$$y''_{p1}(x) + p(x)y'_{p1}(x) + q(x)y_{p1}(x) = g(x)$$ $$y''_{p2}(x) + p(x)y'_{p2}(x) + q(x)y_{p2}(x) = g(x)$$ $$ \Rightarrow (y''_{p1}(x)-y''_{p2}(x)) + p(x)(y'_{p1}(x)-y'_{p2}(x)) + q(x)(y_{p1}(x)-y_{p2}(x)) = 0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.