# Integration (DE)

1. Dec 8, 2014

### basty

1. The problem statement, all variables and given/known data

Solve:

$x \frac{dy}{dx} - 4y = x^6 e^x.$

Solution: Dividing by $x$, we get the standard form

$\frac{dy}{dx} - \frac{4}{x}y = x^5 e^x.$

Hence the integrating factor is

$e^{\int -\frac{4}{x}dx}= e^{-4 \int \frac{1}{x}dx} = e^{-4 \ln x} = e^{\ln x^{-4}} = x^{-4}$

Now we multiply $\frac{dy}{dx} - \frac{4}{x}y = x^5 e^x$ by $x^{-4}$ and rewrite

$x^{-4} \frac{dy}{dx} - 4x^{-5}y = x e^x$ as $\frac{d}{dx}[x^{-4}y] = x e^x$

It follows from integration by parts that the general solution defined on the interval $(0, ∞)$ is

$x^{-4}y = x e^x - e^x + c$

or

$y = x^5e^x - x^4e^x + cx^4.$

My question is, why integrate of $x^{-4} \frac{dy}{dx} - 4x^{-5}y = x e^x$ is $y = x^5e^x - x^4e^x + cx^4$?

Compare it with mine below.

Which one is correct?

2. Relevant equations

3. The attempt at a solution

$x^{-4} \frac{dy}{dx} - 4x^{-5}y = x e^x$

Multiplying by $dx$ yield

$x^{-4} dy - 4x^{-5}y \ dx = x e^x dx$

Integrating both sides

$\int x^{-4} dy = x^{-4}y + c_1$

$\int - 4x^{-5}y \ dx = -4 \int x^{-5}y \ dx = -4 \frac{1}{-4}x^{-4}y + c_2 = x^{-4}y + c_2$

$\int x e^x dx = xe^x - e^x + c_3$

So

$\int x^{-4} dy - \int 4x^{-5}y \ dx = \int x e^x dx$
$x^{-4}y + c_1 + x^{-4}y - c_2 = xe^x - e^x + c_3$
$x^{-4}y + x^{-4}y = xe^x - e^x - c_1 + c_2 + c_3$
$x^{-4}y + x^{-4}y = xe^x - e^x + c_4$
$2x^{-4}y = xe^x - e^x + c_4$
$2y = \frac{xe^x - e^x + c_4}{x^{-4}}$
$2y = x^5e^x - x^4e^x + c_4x^4$
$y = \frac{x^5e^x - x^4e^x + c_4x^4}{2}$

2. Dec 8, 2014

### Staff: Mentor

I think you're asking why the work in the first part of your post differs from the work and the second part.
The short answer is that it (your solution equation) is not the integral of your differential equation. In the work of the first part, the original differential equation was manipulated to get to this equation:
$\frac{d}{dx}[x^{-4}y] = x e^x$
Since you know that the derivative of x-4y is xex, it follows that x-4y equals an antiderivative of xex.

Now if you're asking why the work of the first part produces a different result than the work in the second part, the best I can come up with is that your integrals aren't right. Specifically $\int x^{-4} dy$ and $\int - 4x^{-5}y \ dx$.

When you solve a diff. equation you are tacitly assuming that there is some function y = f(x) that satisfies the diff. equation. In your integrals above, you are treating x and y as being independent of each other, which contradicts the basic assumption that y is a function of x. I'm surprised that the result in the second part came out so close to the solution in the first part.
The work in the first part looks correct. You can check each solution by differentiating to get back to the original diff. equation.

3. Dec 8, 2014

### ehild

Your method is wrong. x and y are not independent variables. While integrating with respect y, you can not consider x a constant.

4. Dec 8, 2014

### basty

How do he obtain

$x^{-4}y = x e^x - e^x + c$

or

$y = x^5e^x - x^4e^x + cx^4.$

?????????????

5. Dec 9, 2014

### Staff: Mentor

The whole point of getting an integrating factor is to be able to write one side of the differential equation as the derivative of something.
After one step in the first part of your post you have
y' - (4/x)y = x5ex

In the later work an integrating factor of x-4 was found.

Multiplying through by the integrating factor yields this equation:
x-4y' - 4x-5y = xex

The left side of the equation above is the derivative of x-4y,
so d/dx(x-4y) = xex
or x-4y = $\int xe^x dx$

Finally, integrate the right hand side, and then multiply both sides by x+4 to get y in terms of x alone.

6. Dec 9, 2014

### basty

How to differentiate

$x^{-4}y$

or

$\frac{d}{dx}(x^{-4}y)$

????

7. Dec 9, 2014

### ehild

How do you differentiate a product? Keep in mind that y is a function of x. You need to get
$\frac{d}{dx}(x^{-4}y(x))$

8. Dec 9, 2014

### Staff: Mentor

How is it that you're taking a differential equations class (or at least studying them) but don't know how to differentiate a product? Differentiating products is one of the first things you learn in a calculus class.