Mathematics Asked by NX37B on October 8, 2020

I came across this question about homogenous differential equations, which has me scratching my head because it seems like its doing magic. The problem is as follows:

A particular solution of the differential equation $xfrac{dv}{dx}=f(v)$ has $v=2$ when $x=e$. If $k$ is the value of $v$ when $x=1$, show that $int_k^2frac{1}{f(y)}dy=1$.

I started by using separation of variables to get that $$intfrac{1}{f(v)}dv=intfrac{1}{x}dx$$

$$intfrac{1}{f(v)}dv=ln(x)$$

$$int_k^2frac{1}{f(v)}dv=left[ln(x)right]_1^e=1$$

Here is when I no longer understand what am I supposed to do. Should I just say that it does not really matter which variable is being used in the function $f(x)$ because it will always be the same? I also don’t really know if the question assumes that $y=vx$, as is used in v-substitution, but it was never explicitly stated.

You have probably confused yourself. First step is correct but the second is not completely precise.

You cannot easily switch like that from indefinite to definite integral. When you use a substitution you have to substitute the interval as well. So this would be correct.

$$int_k^2frac{1}{f(v)}dv = int_{1}^{e}frac{1}{x}dx = left[ln(x)right]_1^e=1$$

You have simply used the substitution $dv=f(v)frac{dx}{x}$.

Answered by Aleksandar Perisic on October 8, 2020

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