# dy/dx = xe^(y-2x), form differntial eqaution [Solved!]

**grabbitmedia** 02 Dec 2016, 01:42

### My question

`dy/dx = xe^(y-2x)` , i am asked to form differential equation using this equation . the ans given is

`e^-y = 0.5(e^(-2x))(x+0.5) + a`

how to get the answer?

### Relevant page

2. Separation of Variables

### What I've done so far

`dy/dx=xe^(y-2x)`

`dy/dx=x(e^y)/(e^(-2x))`

`e^(-y)dy = xe^(-2x)dx`

`int e^(-y)dy = int xe^(-2x)dx`

`-e^(-y) = -1/2int -2xe^(-2x)dx`

`-e^(-y) = -1/2e^(-2x)+C`

X

`dy/dx = xe^(y-2x)` , i am asked to form differential equation using this equation . the ans given is
`e^-y = 0.5(e^(-2x))(x+0.5) + a`
how to get the answer?

Relevant page
<a href="/differential-equations/2-separation-variables.php">2. Separation of Variables</a>
What I've done so far
`dy/dx=xe^(y-2x)`
`dy/dx=x(e^y)/(e^(-2x))`
`e^(-y)dy = xe^(-2x)dx`
`int e^(-y)dy = int xe^(-2x)dx`
`-e^(-y) = -1/2int -2xe^(-2x)dx`
`-e^(-y) = -1/2e^(-2x)+C`

## Re: dy/dx = xe^(y-2x), form differntial eqaution

**Murray** 04 Dec 2016, 04:13

Hello

I don't believe your question says "form a differential equation" since your question is **already** a differential equation! Perhaps the original says "Solve the differential equation"?

Your working for the first 4 lines is correct, but your answer for the integral `int xe^(-2x) dx` is not. (A hint: It's a product, not a straightforward integral.)

This page should help: Integration by Parts. Example 5 is almost identical to the integral you need to do (actually, yours requires less steps).

X

Hello
I don't believe your question says "form a differential equation" since your question is <b>already</b> a differential equation! Perhaps the original says "Solve the differential equation"?
Your working for the first 4 lines is correct, but your answer for the integral `int xe^(-2x) dx` is not. (A hint: It's a product, not a straightforward integral.)
This page should help: <a href="http://www.intmath.com/methods-integration/7-integration-by-parts.php">Integration by Parts</a>. Example 5 is almost identical to the integral you need to do (actually, yours requires less steps).

## Re: dy/dx = xe^(y-2x), form differntial eqaution

**Murray** 31 Jan 2017, 01:06

It seems grabbitmedia has disappeared. Anyone else like to have a go at finishing this?

X

It seems grabbitmedia has disappeared. Anyone else like to have a go at finishing this?

## Re: dy/dx = xe^(y-2x), form differntial eqaution

**stephenB** 01 Feb 2017, 04:53

Using Integration by Parts, put

`u = x` and `dv = e^(-2x)dx`

This gives

`du=dx` and `v=-(e^(-2x))/2`

Then

`int u dv = uv - int vdu`

means

`int xe^(-2x)dx = x(-e^(-2x)/2)-int-(e^(-2x))/2dx`

`=-xe^(-2x)/2-(e^(-2x))/4`

`=-(e^(-2x))(x/2+1/4)+C`

Putting it all together, it's

`-e^(-y) = -(e^(-2x))(x/2+1/4)+C`

`e^(-y)= e^(-2x)(2x+1)/4+C`

X

Using Integration by Parts, put
`u = x` and `dv = e^(-2x)dx`
This gives
`du=dx` and `v=-(e^(-2x))/2`
Then
`int u dv = uv - int vdu`
means
`int xe^(-2x)dx = x(-e^(-2x)/2)-int-(e^(-2x))/2dx`
`=-xe^(-2x)/2-(e^(-2x))/4`
`=-(e^(-2x))(x/2+1/4)+C`
Putting it all together, it's
`-e^(-y) = -(e^(-2x))(x/2+1/4)+C`
`e^(-y)= e^(-2x)(2x+1)/4+C`

## Re: dy/dx = xe^(y-2x), form differntial eqaution

**Murray** 02 Feb 2017, 08:40

Looks good to me, stephenB!

X

Looks good to me, stephenB!

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