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I know from linear algebra that for a non-homogeneous consistent linear system that has an infinite number of solutions, if the last row of the augmented matrix has all zeros, then we have an infinite number of particular solutions.

I was wondering if there is a similar parallel with differential equations, namely say I have a differential equation of the form

$y''(x)(Ax+B)^2 + y(x) =$ some cubic, and I want to find a particular solution of this. Then I make an $ansatz$ that my $y(x)$ is some cubic as well, of the form $y(x) = a_0 x + a_1 x + ... a_3 x^3$

When comparing coeffcients on the left and right hand sides, I have come across cases where by the coeffcients $a_0$, ... $a_1$ on the L.H.S. do not have particular values, but are instead expressed as expressions involving one of the other coefficients as a free variable.

Is it possible that differential equations too may have an infinite number of particular solutions?

Ben

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There is in fact a very strong connection between (certain kinds of) differential equations and systems of linear equations. They are both instances of linear transformations.

Consider the collection of all infinitely differentiable functions. There is a function $D$ from this set to itself that maps a function $f(x)$ to $$D(f) = f''(x)(Ax+B)^2 + f(x).$$ This map is linear: $D(f+g) = D(f)+D(g)$, and $D(\alpha f) = \alpha D(f)$ for all infinitely differentiable functions $f$ and $g$, and all real numbers $\alpha$.

In particular, fix a function $k_0(x)$, and suppose that $D(f) = k_0(x)$ (that is, $f$ is a particular solution to $y''(x)(Ax+B)^2 + y(x) = k_0(x)$). Then:

  1. If $g(x)$ is any solution to $D(g)=0$, that is, $g''(x)(Ax+B)^2 + g(x)=0$, then $f(x)+g(x)$ is also a solution to $D(y)=k_0(x)$. This because $D(f+g) = D(f)+D(g) = k_0(x)+0 = k_0(x)$.

  2. If $h$ is any other solution to $D(y)=k_0(x)$, then there exists a solution $g$ to $D(g)=0$ such that $h(x) = f(x)+g(x)$. That is, all solutions to $D(y)=k_0(x)$ are obtained by taking a single particular solution and addding solutions to $D(y)=0$.

    To see this, notice that if $D(f)=D(h)=k_0(x)$, then $D(h-f) = D(h)-D(f) = k_0(x)+k_0(x) = 0$. So $h-f$ is a solution to $D(y)=0$. And $h=f+(h-f)$, so $h$ can be obtained by adding a solution to $D(y)=0$ to $f$.

In particular, the differential equation $$y''(x)(Ax+B)^2 + y(x) = \text{some cubic}$$ that has at least one particular solution has exactly as many particular solutions as there are solutions to $$y''(x)(Ax+B)^2 + y(x) = 0.$$

So if the homogeneous equation has infinitely many (essentially distinct) solutions, and your original equation has at least one particular solution, then the original equation will in fact have infinitely many (essentially distinct) solutions as well.

This is in complete analogy to the case of a system of linear equations, which can be seen as given by the matrix equation $A\mathbf{x}=\mathbf{b}$, where $A$ is the matrix of coefficients of the system: given any particular solution $\mathbf{x}_0$, you get all solutions by taking $x_0+\mathbf{z}$, where $\mathbf{z}$ is any solution to $A\mathbf{z}=\mathbf{0}$.

This is all part of linear algebra. Both systems of linear equations and linear differential equations can be studied together, and many general results apply to both, so it's no surprise you are seeing a lot of similarities: in fact, there are a lot of similarities.

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Thanks for your reply. So say for example I have an ODE $y''(1+x)^2-2y(x) = (1+x)^3$ I guess that my solution $y_{particular}$ is of the form $Ax^3 + Bx^2+..D$, and so if I sub in $y$ into the ODE above I see that $A = f(D)$, $B = f(D)$, $C = f(D)$, where $D$ is my free variable. I mean if I were to write down the general solution to this ODE, I could only pick of the infinitely many particular solutions to write $y = y_p + y_c$, right? –  user38268 Apr 3 '11 at 2:21
    
@David: I don't know what $y_c$ is; if you mean, a solution to the corresponding homogeneous equation, then yes: if you know one particular solution, $y_p$, then every solution is of the form $y_p + y_h$, where $y_h$ is a solution to the corresponding homogeneous equation. –  Arturo Magidin Apr 3 '11 at 2:51
    
Yep, $y_c$ is standard undergrad Aussie notation for a solution to the corresponding homogeneous DE ('c' for 'complementary'). –  Glen Wheeler Apr 3 '11 at 8:29
    
Hi that makes it clear then, oh sorry I forgot to say what $y_c$ was, I should have said $y_c$ is the linear combination of all the solutions in the null space. –  user38268 Apr 3 '11 at 10:22
    
@David: Well, not "the linear combination of all the solutions in the null space" (there is no such thing as the linear combination, and "all solutions in the null space" is a lot of solutions). But I assume that is a mistake of not saying what you mean, rather than meaning what you said (-; Yes: $y_c$ is a solution in the null space, or a linear combination of the basis solutions in the null space. –  Arturo Magidin Apr 3 '11 at 19:05

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