<s>[INST] First order differential equation solution check. I have an equation:
$\frac{dy}{y}=(0.0001x^2+0.005x)dx$
$y=ce^{(0.00003x+0.0025)x^2}$
Now I try to substitute $y$ in the euqation and I am getting
$(\frac{1}{ce^{(0.00003x+0.0025)x^2}})'=0.0001x^2+0.005x$
$(\frac{1}{ce^{(0.00003x+0.0025)x^2}})'=(\frac{1}{c}e^{(-0.00003x-0.0025)x^2})'=(\frac{1}{c}e^{(-0.00003x-0.0025)x^2})(-0.0001x^2-0.005x)$
But
$(\frac{1}{c}e^{(-0.00003x-0.0025)x^2})(-0.0001x^2-0.005x) \neq 0.0001x^2+0.005x$
What am I missing here while checking the solution?
 [/INST] The equation is of the form: 
$$ \frac{\mathrm{d}y}{y} = (ax^2+bx)\mathrm{d}x $$
where $a=0.0001$ and $b=0.005$. Integrating both sides gives: 
$$ \int\frac{\mathrm{d}y}{y} = \int(ax^2+bx)\mathrm{d}x $$
The solution is
$$ \ln y = \frac{a}{3}x^3+\frac{b}{2}x^2+c$$
Therefore, in terms of $y$:
$$ y = C\mathrm{e}^{(\frac{a}{3}x+\frac{b}{2})x^2}, \label{eq1} \tag{1}$$
where $C=\mathrm{e}^c$. We can test the solution in the following way: 
$$\begin{align} 
\frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}}{\mathrm{d}x}\left [ C\mathrm{e}^{(\frac{a}{3}x+\frac{b}{2})x^2} \right]\\
&=\frac{\mathrm{d}}{\mathrm{d}x}\left [ \frac{a}{3}x^3+\frac{b}{2}x^2 \right]C\mathrm{e}^{(\frac{a}{3}x+\frac{b}{2})x^2} \\
&=\left ( ax^2+bx^2 \right)C\mathrm{e}^{(\frac{a}{3}x+\frac{b}{2})x^2} 
\end{align}.$$ 
Using $\eqref{eq1}$:
$$
\frac{\mathrm{d}y}{\mathrm{d}x} =\left ( ax^2+bx^2 \right)y 
$$
Therefore: 
$$
\frac{\mathrm{d}y}{y} =\left ( ax^2+bx^2 \right) \mathrm{d}x
$$
Regarding the numeric values: $a=0.0001=\frac{1}{10000}$, then $\frac{a}{3}=\frac{1}{30000}\neq 0.00003$. On the other hand, 
$$ 
\frac{\mathrm{d}y}{y} \neq \frac{1}{y'}
$$
The right way to view the equation is the following: The differential of a function $y(x)$ is given by
$$
\mathrm{d}y=y'(x)\mathrm{d}x
$$
Divide each side of the equation by $y(x)$
$$
\frac{\mathrm{d}y}{y}=\frac{y'(x)}{y}\mathrm{d}x.
$$
Then $\frac{y'(x)}{y}$ should give you $0.0001x^2+0.005x$. 
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