Find a function that satisfies the following Differential Eq

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

Homework 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.

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.

Thank you for your help!

 

[ehild]

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

Homework 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.

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.

Thank you for your help!

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

 

[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))

 

[ehild]

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))

It will be all right.

 

[SqueeSpleen]

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

 

[Fred Wright]

All ready answered while I was typing.

 

[ehild]

Also, your signature is great, having at hand those symbols is very convenient.

You find those symbols and a couple of more by hitting the ∑ button above if you write a post. .