+0

# Differential equation - I think I got it right but Alan says its wrong so any help?

0
68
4
+138

Differential equation$$\frac{dy}{dx}=1+\frac{y}{x}$$

Initial Condition (1,1)

My work as follows: First I rewrite as a first order linear

$$y'\:-\frac{1}{x}y=1$$

then find the integrating factor

$$μ\left(x\right)=\frac{1}{x}$$

Then multiply by integrating factor

$$\frac{y'\:}{x}-\frac{y}{x^2}=\frac{1}{x}$$

Then product rule

$$\left(\frac{1}{x}y\right)'\:=\frac{1}{x}$$

And then solve for this

$$y=\ln \left(x\right)x+cx$$

c=1,

so

$$y=\ln \left(x\right)x+x$$

Am I doing something wrong here? Any help is appreciated :)

Mar 12, 2021

### 4+0 Answers

#1
+31281
+1

Maybe you overthought it a bit, but I think it is just this:

1 + y x-1   dy/dx

=   -1 y x-2

= - y/x2                                          ??

Mar 12, 2021
edited by ElectricPavlov  Mar 12, 2021
#2
+138
0

what??? lmao im lost

SpaceTsunaml  Mar 12, 2021
#3
0

I solved the de as you did and got exactly the same result.

You can check that it's correct by differentiating it and seeing that it takes you back to the original de.

So,

$$\displaystyle y = x \ln(x) +x, \\ y' = \ln(x) + x/x + 1 =\ln(x)+2, \\ xy'=x\ln(x)+2x =( x\ln(x)+x)+x=y+x,\\ y'=y/x+1,$$

which is the original de, so with you, I think that it's correct.

Mar 13, 2021
#4
+32062
0

Your result is correct.  I was too hasty!

Mar 13, 2021