Solution for ∫y(x)dx = ky, = k/y, = kx, = k/x

  • Thread starter Thread starter Jhenrique
  • Start date Start date
Jhenrique
Messages
676
Reaction score
4

Homework Statement



Get the solution (for y(x)) for the follows integrals: \int y(x) dx = ky\;\;\;\;\;(1) \int y(x) dx = \frac{k}{y}\;\;\;\;\;(2) \int y(x) dx = kx\;\;\;\;\;(3) \int y(x) dx = \frac{k}{x}\;\;\;\;\;(4)

Homework Equations



The Attempt at a Solution



\\ \int y(x)dx = kx \\ \\ \int y(x)\frac{dx}{dx} = \frac{kx}{dx} \\ \\ d\int y(x) = d\frac{kx}{dx} \\ \\ y(x) = k\frac{dx}{dx} \\ \\ y(x) = k

\\ \int y(x)dx = \frac{k}{x} \\ \\ \int y(x)\frac{dx}{dx} = \frac{1}{dx} \frac{k}{x} \\ \\ d\int y(x) = \frac{d}{dx} \frac{k}{x} \\ \\ y(x) = k \frac{d}{dx}\left ( \frac{1}{x} \right ) \\ \\ y(x) = -\frac{k}{x^2}
 
Physics news on Phys.org
Jhenrique said:

Homework Statement



Get the solution (for y(x)) for the follows integrals: \int y(x) dx = ky\;\;\;\;\;(1) \int y(x) dx = \frac{k}{y}\;\;\;\;\;(2) \int y(x) dx = kx\;\;\;\;\;(3) \int y(x) dx = \frac{k}{x}\;\;\;\;\;(4)

Homework Equations



The Attempt at a Solution



\\ \int y(x)dx = kx \\ \\ \int y(x)\frac{dx}{dx} = \frac{kx}{dx} \\ \\ d\int y(x) = d\frac{kx}{dx} \\ \\ y(x) = k\frac{dx}{dx} \\ \\ y(x) = k

\\ \int y(x)dx = \frac{k}{x} \\ \\ \int y(x)\frac{dx}{dx} = \frac{1}{dx} \frac{k}{x} \\ \\ d\int y(x) = \frac{d}{dx} \frac{k}{x} \\ \\ y(x) = k \frac{d}{dx}\left ( \frac{1}{x} \right ) \\ \\ y(x) = -\frac{k}{x^2}

I'm not sure anyone would buy your reasoning, especially dividing by dx that you have done.
For problem 3, ∫y(x)dx = kx, differentiate both sides with respect to x to get y(x) = k.
For problem 4, ∫y(x)dx = k/x, differentiate both sides with respect to x to get y(x) = -k/x2.
For problem 1, which you didn't try, ∫y(x)dx = ky. As before, differentiate both sides with respect to x to get y(x) = k dy/dx. This is a separable differential equation that is pretty easy to solve.
 
  • Like
Likes 1 person
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top