General solution of differential equation (express y in term of x)

[delsoo]

Homework Statement

i got stucked here. below is the answer given. can anybody help please?

Homework Equations

The Attempt at a Solution

 

[haruspex]

You can try to simplify the original equation by substituting y(x) = f(x).xⁿ, then see what value of n will get rid of the -y/x term.

 

[Chestermiller]

This looks like a straightforward "integrating factor" problem.

Chet

 

[delsoo]

dy/dx +Py(x) = Q(X) if i rearrange i would get 0.5 dy/dx + y/x = arc tan x ... is arc tan x function of x?

 

[Chestermiller]

dy/dx +Py(x) = Q(X) if i rearrange i would get 0.5 dy/dx + y/x = arc tan x ... is arc tan x function of x?

Multiply both sides by 2. arc tan x is a function of x.

Chet

 

[delsoo]

i redo the question and don't know how to proceed here... any idea on how should i do next ? i don't know how to integreate arc tan x

 

[Chestermiller]

i redo the question and don't know how to proceed here... any idea on how should i do next ? i don't know how to integreate arc tan x

Integrate by parts. Do you remember how to take the derivative of arc tan x with respect to x?

Chet

 

[delsoo]

sorry , i checked thru the syllabus, there's no deriative of arc tan x in it or maybe i can do it in other way? any other way?

 

[Chestermiller]

sorry , i checked thru the syllabus, there's no deriative of arc tan x in it or maybe i can do it in other way? any other way?

Yes. Let y = arc tan x

Then tan y = x

Differentiating both sides with respect to x;

[tex]sec^2y\frac{dy}{dx}=1[/tex]

Also, we have the trig identity: [itex]tan^2y+1=sec^2y[/itex]

So, [itex]sec^2y=1+x^2[/itex]

So, [tex]\frac{dy}{dx}=\frac{1}{1+x^2}[/tex]

So, [tex]\frac{d(tan^{-1}x)}{dx}=\frac{1}{1+x^2}[/tex]

Chet

 

[delsoo]

thanks got the solution finally!