I am given:
\begin{equation}xy'=2y\end{equation}
Therefore,
\begin{equation}y'=(2y)/x\end{equation}
Solving for y, I have:
$y=Cx^2$ (Originally, was $e^cx^2$, but C is arbitrary, so I simplified)
To start verifying the solution, I inserted y' into the original equation:
\begin{equation}x(\frac{2y}{x})=2y\end{equation}
Next, I substitute the value of y: \begin{equation}x(\frac{2Cx^2}{x})=2(Cx^2)\end{equation}
After simplifying both sides, I have:
\begin{equation}2Cx^2=2Cx^2\end{equation}
Did I do this right?
Thanks!
$\endgroup$ 14 Answers
$\begingroup$Right solution, wrong verification.
you have to get $y'=2Cx$ from your final solution and then substitute to get the same result.
We have $y=Cx^2$. That means $y'=2Cx$. Let's plug those in to your original differential equation $$xy'=2y.$$ That is $$x(2Cx)=2(Cx^2)$$ which reduces to $$2Cx^2=2Cx^2$$ which verifies that you have the correct solution.
$\endgroup$ $\begingroup$What you "verified" is that, if $y'=2y/x$ and if $y=Cx^2$ then $xy'=2y$. Since the condition $y'=2y/x$ alone implies that $xy'=2y$ this is not a verification at all.
Rather, assume that $y=Cx^2$. Then $y'=2Cx$ hence $xy'-2y=x(2Cx)-2(Cx^2)=0$, that is, $xy'=2y$.
To sum up, one is supposed to check that, if $y=Cx^2$ then $y$ solves the equation $xy'=2y$.
$\endgroup$ $\begingroup$You need to take the derivative of y first before you can verify your equation.
$y = Cx^2$
$y' = C(2x) +x^2(0) = 2xC$
Recall that your equation is
$\begin{equation}xy'=2y\end{equation}$
now substitute the result from $y'$ and the original $y$
$\begin{equation}xy'=2y\end{equation}$
$\begin{equation}x(2xC)=2(Cx^2)\end{equation}$
$\begin{equation}(2x^2C)=2(Cx^2)\end{equation}$
$\endgroup$
