Suppose a, b, c are three real numbers such that

[nysnacc]

Homework Statement

Homework Equations

character equation

The Attempt at a Solution

Should I set a = ax² b= bx c =c in the character equation?

 

[Ssnow]

Hi, you must start to substitute ##y_{1}(x)=x^{r_{1}}## in the differential equation and find ##r_{1}## that satisfy the equation ...,
so start to calculate ##y'_{1}(x)## and ##y_{1}''(x)## ...

The same for ##y_{2}(x)=x^{r_{2}}\ln{x}## ...

 

[nysnacc]

y₁ = x^r1
y₁' = r1* x^r1-1
y₁" = r1 (r1-1)* x^r1-2 = ( r1² - r1) *x^r1-2

then plug in ODE?

 

[nysnacc]

Then (I simply say r instead of r1)
a(r²-r)x^r +b r x^r + cx^r

x^r ( a(r²-r) + b r +c )
And solve for r ?

 

[Mark44]

Then (I simply say r instead of r1)
a(r²-r)x^r +b r x^r + cx^r

x^r ( a(r²-r) + b r +c )
And solve for r ?

You can't solve the above, because they aren't equations!
The equation you need to solve is a(r²-r)x^r +b r x^r + cx^r = 0,
which is equivalent to x^r ( a(r²-r) + b r +c ) = 0
Now solve for r.

 

[nysnacc]

Oh yeah! but I don't know a, b and c... there are four unknowns then

 

[Mark44]

Oh yeah! but I don't know a, b and c... there are four unknowns then

But you have a relationship involving a, b, and c given in your problem statement.

 

[nysnacc]

So I solve for r which is r = ... (in terms of a, b, c) probably be -b +/- sqrt (b^2 - 4 ac) / 2a

Then use the relationship involving a, b, and c given in problem statement?

But do I need one more equation?

 

[nysnacc]

For r1, it is -B/2A which is -(b-a)/2a
Do I end up wth the letter as coefficient, or any further?

 

[Mark44]

So I solve for r which is r = ... (in terms of a, b, c) probably be -b +/- sqrt (b^2 - 4 ac) / 2a

No it isn't.

Start from this:
x^r ( a(r²-r) + b r +c ) = 0
The part on the left has to be identically equal to zero; i.e., for all values of x.

So either x^r = 0 (can't be true)
or a(r²-r) + b r +c = 0
Solve this equation for r.

Then use the relationship involving a, b, and c given in problem statement?

But do I need one more equation?

 

[Mark44]

For r1, it is -B/2A which is -(b-a)/2a
Do I end up wth the letter as coefficient, or any further?

Yes, this is what I get: r = -(b - a)/(2a), which is the same as (a - b)/(2a)

 

[nysnacc]

Great, so this is it, for r1 ?? :)

 

[Mark44]

Great, so this is it, for r1 ?? :)

Yes.

 

[nysnacc]

For r2,
I have y2 = x^r2 ln (x)
y2' = x^r2-1 +r2 x^r2-1 ln (x)
y2" = (r2-1)x^r2-2 +r2 x^r2-2 + (r2^2-r2)x^r2-2 ln(x)

Which then plug into the ODE, i found r2 = -(b-a)/ 2a same as r1

and whole ODE can be 0 if x =1