Differential partial equation HELP

[AnnSte]

Hello!
I would appreciate if someone could lead me to find the solution for this:
What is y(2) if y(0)=4 for y'-(4x/y)=4.
Thank you in advance.

 

[MarkFL]

Hello, and welcome to FMH!

I would write the ODE as:

[MATH]\frac{dy}{dx}=4\frac{x}{y}+4[/MATH]
Next, let's consider the substitution:

[MATH]v=\frac{y}{x}\implies y=vx\implies \frac{dy}{dx}=\frac{dv}{dx}x+v[/MATH]
And now our ODE becomes:

[MATH]\frac{dv}{dx}x+v=\frac{4}{v}+4[/MATH]
[MATH]\frac{dv}{dx}x=\frac{4+4v-v^2}{v}[/MATH]
Can you proceed with this now separable equation?

 

[AnnSte]

Thank you so far! I can solve the right part but unfortunately I don’t understand what to do with the left part after that
I will try to solve it tomorrow morning and will return if I cannot.
Thank you again!

 

[MarkFL]

Separating the variables, we can write:

[MATH]-\frac{v}{v^2-4v-4}\,dv=\frac{1}{x}\,dx[/MATH]
Now, let's examine the denominator on the LHS:

[MATH]v^2-4v-4[/MATH]
By the quadratic formula, it has the roots:

[MATH]v=2\pm2\sqrt{2}[/MATH]
And so we can write:

[MATH]-\frac{v}{(v-(2+2\sqrt{2}))(v-(2-2\sqrt{2}))}\,dv=\frac{1}{x}\,dx[/MATH]
At this point, I would consider a partial fraction decomposition on the LHS.

 

[Deleted member 4993]

Alternatively,

continuing from response #2, I would note that

v^2 - 4v - 4 = (v-2)^2 - 8

substituting

u = v-2, we would get,

(u+2)/(u^2 - 8) du

= u/(u^2 - 8) du + 2/(u^2 - 8) du

These are almost standard integrals.