Show that $\int_{k}^{2} \frac{1}{f(y)} dy = 1$ Let us have the differential equation $x\frac{dv}{dx}=f(v(x))$ and the solution $v$ s.t. $v(e)=2$. Let $v(1)=k$. Show that $$\int_{k}^{2} \frac{dy}{f(y)} = 1.$$
Can anyone please help me understand how to show that?
I was taught that $v$ is defined as $\frac yx$. I understand how to show that but how is it done if we assume that $v=\frac yx$?
 A: Rearrange and get
$$\frac{dx}{x} = \frac{dv}{f(v)} \implies \int_1^e\frac{dx}{x} = \int_k^2\frac{dv}{f(v)}  \implies\ln(e) - \ln(1) = \int^2_k\frac{dv}{f(v)}$$
Let $v=y$
$$\implies 1-0 = 1 =\int^2_k\frac{dy}{f(y)}$$
A: Substitute directly for $f$.  Use the inverse function theorem to rewrite the reciprocal of the derivative outside of the fraction.  Use the chain rule (in reverse of the way you used it repeatedly for implicit differentiation) to simplify the product of derivative and the differential element.  Then use the data about the differential equation's particular values to change the endpoints of integration.  Now you have an elementary integral; evaluate it.

 \begin{align*}  \int_{k}^{2}\; \frac{1}{f(y)} \,\mathrm{d}y  &= \int_{k}^{2}\; \frac{1}{x \frac{\mathrm{d}y}{\mathrm{d}x}} \,\mathrm{d}y  \\      &= \int_{k}^{2}\; \frac{1}{x} \frac{\mathrm{d}x}{\mathrm{d}y} \,\mathrm{d}y  \\      &= \int_{y = k}^{y = 2}\; \frac{1}{x}  \,\mathrm{d}x  \\      &= \int_{1}^{\mathrm{e}}\; \frac{1}{x}  \,\mathrm{d}x  \\    &= \left. \ln x \right|_1^{\mathrm{e}}  \\    &= 1-0  \text{.}  \end{align*}

