# Find a function that satisfies the following Differential Eq

1. Jun 17, 2017

### SqueeSpleen

I'm helping some guys with Calculus I class and found this exercise in the practice about integrals.
I think it's overkill but it may have some easy way to solve it.
I'm very rusty solving differential equations.

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

Find f differentiable such that
$$(3+f'(x))e^{2-x} = (x-6) (3x+f(x))^{2}$$
with f(2)=0.

2. Relevant equations
I have solved similar exercises searching a function g(x) such that g(f(x))'=g'(f(x)).f'(x) and and putting the equation in a way that you have g'(f(x))=h(x). So you can integrate and solve it. The most classic example is (ln(f(x))'=f'(x)/f(x), but I have done with more complicated functions. In this case I had to go to wolframalpha because I couldn't figure out the solution, and even seeing the solution I'm not having an easy time finding it, I only was able to verify that's a solution. If it were a linear differential equation it would be easier, but the square spoils everything.

3. The attempt at a solution
I also thought about integrating by parts the left part of the equality, but I'm not sure if it will help to arrive an easier equation as I don't know what to do with the expressión on the right.

Any suggestions? I'm very rusty with differential equations but as this exercise was in a calculus 1 class I think it should be easier to solve.

2. Jun 17, 2017

### ehild

Try to solve it for the function g(x) = 3x+f(x)

3. Jun 17, 2017

### SqueeSpleen

Let's see if I got it.
Let g(x)=f(x)+3x. Then g'(x)=f'(x)+3.
So the equation becomes
$$g'(x).e^{2-x} = (x-6) g(x)^{2}$$
We work the expression
$$\dfrac{g'(x)}{ g(x)^{2} } = (x-6) e^{x-2}$$
We re-write the expression on the left
$$-( \dfrac{1}{g(x)} )' = \dfrac{g'(x)}{ g(x)^{2} } = (x-6) e^{x-2}$$
So
$$( \dfrac{1}{g(x)} )' = (6-x) e^{x-2}$$
Now all we have to do is to integrate and find the g(x) (and so f(x))

4. Jun 17, 2017

### ehild

It will be all right.

5. Jun 17, 2017

### SqueeSpleen

Thank you for your help. Also, your signature is great, having at hand those symbols is very convenient.

6. Jun 17, 2017