Formation of differential equations if solutions are known Can someone give me hints as to how am i to form a differential equation if i know the solutions of the differential equation. As for instance if $x^2$ and $x^2\log x$ are the solutions, then how will i determine the differential equation? (The solutions are linearly independent.)
 A: The generalization of the linear version is that
$$\sum_{k=0}^{n-1}c_k(x)y_j^{(k)}(x)+y_j^{(n)}(x)=0$$
Is the $n^{th}$-order linear homogeneous differential equation the functions $y_j(x)$, $1\le j\le n$ must satisfy. This can be rewritten as
$$W^T(x)\vec c(x)=-\vec Y_n(x)$$
Where $W(x)$ is the Wronskian matrix,
$$W_{kj}(x)=\frac{d^k}{dx^k}\left(y_j(x)\right)$$
And $\vec Y_n(x)$ is the column vector with $j^{th}$ element
$$\frac{d^n}{dx^n}\left(y_j(x)\right)$$
For this problem you get $y_1(x)=x^2$, $y_2(x)=x^2\ln x$,
$$W(x)=\begin{bmatrix}x^2&x^2\ln x\\2x&2x\ln x+x\end{bmatrix}$$
$$\vec Y_n(x)=\begin{bmatrix}2\\2\ln x+3\end{bmatrix}$$
So the equation for the coefficients is
$$\begin{bmatrix}x^2&2x\\x^2\ln x&2x\ln x+3\end{bmatrix}\begin{bmatrix}c_0\\c_1\end{bmatrix}=\begin{bmatrix}-2\\-2\ln x-3\end{bmatrix}$$
Taking inverses,
$$\begin{bmatrix}c_0\\c_1\end{bmatrix}=\frac1{2x^3\ln x+x^3-2x^3\ln x}\begin{bmatrix}2x\ln x+x&-2x\\-x^3\ln x&x^2\end{bmatrix}\begin{bmatrix}-2\\-2\ln x-3\end{bmatrix}=\begin{bmatrix}\frac4{x^2}\\-\frac3x\end{bmatrix}$$
So the linear differential equation is
$$y^{\prime\prime}(x)-\frac3xy^{\prime}(x)+\frac4{x^2}y(x)=0$$
If a nonlinear differential equation were acceptable there are several possibilities. For example if $y=x^2(1+c(\ln x-1))$ we get the two solutions when $c=0$ or $c=1$. Then
$$\frac{\frac y{x^2}-1}{\ln x-1}=c$$
Then the numerator of the derivative is
$$\left(\frac{y^{\prime}}{x^2}-\frac{2y}{x^2}\right)(\ln x-1)-\left(\frac y{x^2}-1\right)\left(\frac1x\right)=0$$
So we get to
$$xy^{\prime}(\ln x-1)-2y\ln x+y+x^2=0$$
Which is also satisfied by $y_1(x)$ and $y_2(x)$.
A: If a second-order homogeneous equation with variable coefficients fits your desires, you could use Abel's identity. Say your two functions are $y_1(x)$ and $y_2(x)$, and let the equation you want them to satisfy be $y''+p(x)y'+q(x)=0$ (so our goal is to find $p$ and $q$).
Then (one form of) Abel's identity says that
$$
p(x)=-\frac{W'(x)}{W(x)}
$$
where $W(x)$ is the Wronskian $y_1(x)y_2'(x)-y_1'(x)y_2(x)$. 
Once you know $p(x)$, plug everything back into the original equation to find $q(x)$:
$$
q(x)=-y_1''(x)-p(x)y_1'(x)
$$
and you're done.
A: The solution can be expressed as the form
$$y(x) = Ax^2 + Bx^2\log x$$
for constants $A$ and $B$ (choose $(A,B) = (1,0)$ or $(0,1)$ to recover the given solutions). 
Since there are 2 independent solutions the ODE should be 2nd order. So we find out $y'$ and $y''$:
\begin{align}
y'(x) &= (2A+B)x+2Bx\log x \\
y''(x) &= (2A+3B)+2B\log x.
\end{align}
suppose the ODE is linear i.e.
$$ y'' + p(x)y' + q(x)y + r(x) = 0 $$
note that $p,q,r$ should not involve $A$ and $B$. Therefore we substitute in $y,y',y''$ and collect the coefficients by $A$ and $B$:
$$
A\bigl(2+2px+qx^2\bigr)+B\bigl(3+px+(2+2px+qx^2)\log x\bigr)+r=0
$$
This identity has to hold for all $A$ and $B$, so this can be turned into 3 equations:
\begin{align}
2+2px+qx^2 &= 0 \\
3+px+(2+2px+qx^2)\log x &= 0 \\
r &= 0 
\end{align}
from here it should be easy to solve for $p,q,r$ and gives the original ODE.

 $x^2y'' - 3xy' + 4y=0$

