Separation of variables Calculus The given differential equation I need to solve is $dy/dx=1/x$ with the initial conditon of $x=1$ and $y=10$
My attempt: 
$dy=\frac 1x dx$
Integrating yields 
\begin{align*} 
   y&=\log x+C\\
   x=1 & y=10\\
 10 &=\log 1 + C\\
  10=C\\
\leadsto y&=\log x+10
\end{align*}
Something seems off with my work, if anyone could give me a clue where I went wrong. Thanks
 A: We have $$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{x}$$ Hence, separating the variables gives us $$\int \mathrm{d}y = \ln x + c \implies y=\ln x + c.$$ Using the initial condition of $x=1, y=10$ gives us $$10 = 0 + c \iff c = 10.$$ So the solution is $$\bbox[10px, border: solid lightblue 2px]{y = \ln x + 10}$$ which is exactly what you got and is correct! 

To verify your answer, you can see that when $x=1$, you get $y=10$ and differentiating your answer results in $$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{x},$$ which is what you were given at the beginning. 
A: Your work is fine.  Good job. You might consider an alternative solution: $$\frac{{\rm d}y}{{\rm d}x} = \frac{1}{x} \implies \int \frac{{\rm d}y}{{\rm d}x} \,{\rm d}x = \int \frac{1}{x}\,{\rm d}x \implies y = \ln x + C, $$and now we proceed as you did to get $C = 10 $. My point is: it works, but what does it mean to multiply thorugh ${\rm d}x$, etc?
A: Why do you think you are wrong? The derivative of $\ln(x)$ is in fact $1/x$. You are also satisfying the initial condition $y(1)=10$.
Also there is no need to use separation of variables. We can see directly that $$\frac{dy}{dx} = y'(x) = \frac{1}{x} \implies y(x) = \ln(x) + C.$$
Just as with elementary calculus and algebra, if you are not sure about your answer, then take a derivative and plug it back into the original equation.
A: Sometimes definite integrals help:
$$
    \int_{10}^{y}dy' = \int_{1}^{x}\frac{1}{x'}dx \\
            y-10=\ln x -\ln 1 \\
            y = 10 + \ln x.
$$
