# Ordinary Differential Equations with Arbitrary Constants

Find a differential equation that $y$ satisfies. The ODE should not contain any arbitrary constants. $$y(x)=C_1(x^2-1)+C_2x$$ I proceeded with finding the two derivatives (since it is second-order): $$y'(x)=2C_1x+C_2$$ $$y''(x)=2C_1$$

Then, using the general formula $y''+Ay'+By = 2C_1+A2C_1x+AC_2+BC_1x^2-BC_1+BC_2x$ Which results in $C_1(2+2Ax+Bx^2-B)+C_2(A+Bx)$

Solving the systems of equations, I found that $A=-2x$, $B=2$. Which means the ODE required is $$y''-2xy'+2y=0$$

Can anyone kindly advise if this approach is correct? Thanks.

Edit: Sorry, I found a computational error afterwards. Hence, $$A=-\frac{2x}{x^2+1}$$ $$B=\frac{2}{x^2+1}$$

And it would then arrive at the recommended solution below.

• I'm not exactly sure what it required here, since $y$ is a 2nd degree polynomial, it must satisfy $y''' = 0$, which has no arbitrary constants. Feb 16, 2014 at 22:32
• Also, assuming my computations are correct, $y$ doesn't satisfy the differential equation above, $y''(x)-2x y'(x)+2y(x) = -2 C_1 x^2$. Feb 16, 2014 at 22:42

No. I think what is required is to find the expressions for the integrals of motion $C_1$, $C_2$ in terms of $y$, $y'$ and then substitute them into the expression for $y''$. This gives the equation $$y''=2\frac{xy'-y}{x^2+1}.\tag{1}$$ This is the ''minimal'' equation satisfied by $y(x)$. Of course there are more equations that will be consequences of this one. For example, differentiating it w.r.t. $x$ and substituting $y''(x)$ on the right by (1), you will get the equation $y'''=0$, which was proposed in the comment of @copper.hat.
No, this approach is wrong. Constants are, always, constants. So they do not contain the variable $x$. You need to do back-substitution of the derivative expressions.
$y'(x)=2C_1x+C_2=y''(x)x+C_2$, so $C_1=\tfrac12y''(x)$ and $C_2=y'(x)-xy''(x)$, and thus finally
$y(x)=\tfrac12y''(x)(x^2-1)+(y'(x)-xy''(x))x$.