0
$\begingroup$

Second Order Non-Homo Linear ODE has the general solution of $$y=C_1y_1(t) + C_2y_2(t) +Y(t)$$ where $y_1$ and $y_2$ are the fundamental set of solutions of the corresponding homogeneous ODE.

I'm thinking since $Y$ is not a linear combination of $y_1$ and $y_2$, there should be a coefficient in front as well to make it look more generic. But why does this particular term $Y$ in Non-homogeneous ODE does NOT have a coefficient?

$\endgroup$
  • $\begingroup$ Welcome to MSE. Please use MathJax in the whole question. $\endgroup$ – José Carlos Santos Feb 9 '18 at 19:06
  • 1
    $\begingroup$ $Y(t)$ is just generic function of $t$. Why don't we write $cf(t)$ for a generic function? $\endgroup$ – saulspatz Feb 9 '18 at 19:06
1
$\begingroup$

You have learnt in vectorial geometry of ${\mathbb R}^3$ that a plane $\pi$ in ${\mathbb R}^3$ can be presented in the form $$\pi:\quad (u,v)\mapsto {\bf a}+ u{\bf p}+v{\bf q}\ ,$$ whereby the two vectors ${\bf p}$ and ${\bf q}$ have to be linearly independent. But none of ${\bf a}$, ${\bf p}$, and ${\bf q}$ is uniquely determined by $\pi$. E.g., you could always replace ${\bf a}$ by ${\bf a}'={\bf a}+c_1{\bf p}+c_2{\bf q}$ with arbitrarily chosen factors $c_1$, $c_2$.

Now the solution space of your given ODE is such a plane in an infinite dimensional space of functions. To parametrize this plane we need an initial point ${\bf a}$, called $Y(t)$, and two linearly independent spanning vectors ${\bf p}$ and ${\bf q}$, called $y_1(t)$ and $y_2(t)$ in the ODE context.

$\endgroup$
  • $\begingroup$ This makes a lot sense. Thanks!!! $\endgroup$ – Schwinn Zhang Feb 10 '18 at 17:53
0
$\begingroup$

Let's look at a simple example: $$y''-3y'+2y=1$$ The particular part in this example is $1\over2$ and the homogeneous part is $e^{2t}$ and $e^t$.

We put $C_{1,2}$ to get all the possible solutions for the homogeneous part: $$[C_1e^{2t}+C_2e^t]''-3[C_1e^{2t}+C_2e^t]'+2[C_1e^{2t}+C_2e^t]\\=[C_1e^{2t}]''+[C_2e^t]''-3[C_1e^{2t}]'-3[C_2e^t]'+2[C_1e^{2t}]+2[C_2e^t]\\=C_1(\overbrace{[e^{2t}]''-3[e^{2t}]'+2[e^{2t}]}^0)+C_2(\overbrace{[e^t]''-3[e^t]'+2[e^t]}^0)\\=0$$ But the particular part need to solve the equation, if we add a constant we get:$$[K\frac12]''-3[K\frac12]'+2K\frac12\\=0-3\times0+K\\=K\ne1$$So plugging into the ODE $y=C_1e^{2t}+C_2e^t+K\frac12$ will give you the wrong answer, but $y=C_1e^{2t}+C_2e^t+\frac12$ will.


It is the same in the general case:

for the equation $a(x)y''+b(x)y'+c(x)y=g(x)$ we have $$a(x)[C_1y_1+C_2y_2]''+b(x)[C_1y_1+C_2y_2]'+c(x)[C_1y_1+C_2y_2]\\=a(x)[C_1y_1]''+a(x)[C_2y_2]''+b(x)[C_1y_1]'+b(x)[C_2y_2]'+c(x)[C_1y_1]+c(x)[C_2y_2]\\=C_1(\overbrace{a(x)[y_1]''+b(x)[y_1]'+c(x)[y_1]}^0)+C_2(\overbrace{a(x)[y_2]''+b(x)[y_2]'+c(x)[y_2]}^0)\\=0$$ and for th particular part$$a(x)[KY]''+b(x)[KY]'+c(x)KY\\=K(a(x)[Y]''+b(x)[Y]'+c(x)Y)\\=Kg(x)\ne g(x)$$

$\endgroup$

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.