# Why does the particular term in Non-homogeneous ODE does NOT have a coefficient?

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?

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

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.

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

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$$a(x)[KY]''+b(x)[KY]'+c(x)KY\\=K(a(x)[Y]''+b(x)[Y]'+c(x)Y)\\=Kg(x)\ne g(x)$$