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

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In summary, the provided integrals can be solved by differentiating both sides with respect to x and then solving the resulting differential equations. The solutions are y(x) = k for the first integral, y(x) = -k/x^2 for the second integral, and y(x) = k for the third integral.
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Jhenrique
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Homework Statement



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

Homework Equations



The Attempt at a Solution



[tex]\\ \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[/tex]

[tex]\\ \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}[/tex]
 
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Jhenrique said:

Homework Statement



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

Homework Equations



The Attempt at a Solution



[tex]\\ \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[/tex]

[tex]\\ \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}[/tex]

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

1. What is the solution for ∫y(x)dx?

The solution for ∫y(x)dx is the indefinite integral of the function y(x) with respect to the variable x. It represents the family of all possible functions that have y(x) as their derivative.

2. How do you solve for k in the equation ky?

To solve for k in the equation ky, you need to divide both sides of the equation by y. This will result in k = ky/y, which simplifies to k = 1. Therefore, the solution for k is 1.

3. Can the solution for ∫y(x)dx be expressed in terms of k/y?

Yes, the solution for ∫y(x)dx can be expressed in terms of k/y. This is because when taking the indefinite integral of y(x), the constant of integration can be represented as k/y instead of just k.

4. What is the relationship between ∫y(x)dx and kx?

The relationship between ∫y(x)dx and kx is that they both represent the same function, which is the antiderivative of y(x). The only difference is that kx represents the specific solution for a given value of k, while ∫y(x)dx represents the general solution for all possible values of k.

5. How does the solution for ∫y(x)dx change when k is a negative value?

The solution for ∫y(x)dx remains the same even when k is a negative value. This is because taking the indefinite integral is an operation that is independent of the value of k. However, when using the specific solution kx, a negative value for k will result in a reflection of the graph across the y-axis.

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