Take a linear, non-homogenous ODE:

$$ Ay^{\prime\prime}(x) + By^\prime(x) + Cy(x) = f(x) $$

It is known that one can find the general solution to this equation by taking a solution to the following:

$$ Ay_h^{\prime\prime}(x) + By_h^\prime(x) + Cy_h(x) = 0 $$

And a particular solution of the original ODE:

$$ Ay_p^{\prime\prime}(x) + By_p^\prime(x) + Cy_p(x) = f(x) $$

By adding the two together, you get the following:

$$ A\Big(y_p^{\prime\prime}(x) + y_c^{\prime\prime}(x)\Big) + B\Big(y_p^\prime(x) + y_c^\prime(x)\Big) + C\Big(y_p(x) + y_c(x)\Big) = f(x) $$

Thus, it is proven that a solution to the ODE is given by $y_p(x) + y_c(x)$. Why, though, is this the most general solution? I'm aware the solution is made more general by the introduction of the general homogenous solution, but how do we know that there isn't a more comprehensive solution?

  • $\begingroup$ This is because the LHS of the ODE is linear in $y$. If $y_1$ and $y_2$ are two solutions of the non-homogeneous ODE, their difference $y_1 - y_2$ will be a solution of the homogeneous ODE. $\endgroup$ – achille hui May 6 '19 at 0:18
  • $\begingroup$ I've wondered this as well. Can you clarify @achillehui? Are you saying that because the difference of any two solutions is a solution there will be no other solutions because you just subtract them out? $\endgroup$ – Michael Stachowsky May 6 '19 at 0:34
  • $\begingroup$ Err... OP, you are aware that technically the ODE presented in the first line simplifies to just $Dy = f$ if you define $D := A+B+C$ and factor, right? And similar for later equations. I imagine you meant to include derivatives in there somewhere. $\endgroup$ – Eevee Trainer May 6 '19 at 0:40
  • $\begingroup$ @EeveeTrainer Haha, thanks for pointing that out - fixed. $\endgroup$ – VortixDev May 6 '19 at 0:44

To build on achille hui's point in the comments, this is to do with a larger property of linear maps. If you have a linear map $T : V \to W$, a point $w \in W$, and wish to solve $$Tx = w \tag{1}$$ for $x \in V$, then assuming the equation has a solution $x_p$, then the general solution to $(1)$ is given by \begin{align*} x &= x_p + x_c &\text{where } Tx_c = 0. \end{align*}

To show this property, consider a particular solution $x_p$ to $(1)$. If $x$ is another solution $x$ to $(1)$, then by linearity, $$T(x - x_p) = Tx - Tx_p = w - w = 0,$$ hence there must exist some $x_c$ satisfying $Tx_c = 0$ such that $x - x_p = x_c$, or equivalently, $x = x_p + x_c$. Conversely, if $x = x_p + x_c$, where $x_c$ satisfies $Tx_c = 0$, then by linearity, $$Tx = T(x_p + x_c) = Tx_p + Tx_c = w + 0 = w,$$ hence $x$ satisfies $(1)$ too. That is, $x$ satisfies $(1)$ if and only if it can be expressed as $x = x_p + x_c$ where $Tx_c = 0$, proving the property.

When we have a second order (or any order) linear ODE $$f(x)y'' + g(x)y' + h(x)y = r(x), \tag{2}$$ then we can define a linear operator $T$ from the vector space of twice differentiable functions in some neighbourhood of the initial point, to the space of real-valued functions on the aforementioned neighbourhood by $$T(y) = f \cdot y'' + g \cdot y' + h \cdot y$$ Then $(2)$ reduces to $T(y) = r$, which makes it analogous to $(1)$.

This property of linear maps also can be applied to matrix equations; the general solution to $A\vec{x} = \vec{b}$ is the general solution to $A\vec{x} = \vec{0}$ (aka the nullspace of $A$), but offset by a particular solution $x_p$.


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