solve the differential equation

Question

4x²y"(x)+8xy'(x)+y(x)=0 x is greater than 0; subject to y(1)=2 and y'(1)=1 

a. solve the d.e
b. determine the values of the parameters from the initial condition

Answer 1

This type of equation is called Euler Equations: ax^2y"(x)+bxy'(x)+cy(x)=0

 

in the equation given above, a=4, b=8, c=1.

 

The solution is of the from y(x)= x^r

 

r is the parameter.  

 

y'(x)=r. x^(r-1)

y"(x)= r.(r-1).x^(r-2)

 

let's substitute them into the equation:

 

4x^2.r.(r-1).x^(r-2) +8x.r. x^(r-1) +  x^r =0

4r(r-1)x^r+8rx^r+ x^r =0

x^r(4r^2-4r+8r+1)=0

 

we know that x>0 so x^r cannot be zero. Therefore, so this will only be zero if,

4r^2-4r+8r+1=0

4r^2+4r+1=0

(2r+1)(2r+1)=0

 

r₁= -1/2, r₂=-1/2 (this is double root)

 

the final solution is y(x)= c₁x^r₁ +c₂x^r₂Inx = c₁x^(-1/2) +c₂x^(-1/2)Inx

 

Now using initial condition, we can determine the values of c₁ and c₂:

 

y(1)=2 and y'(1)=1

 

 

y(1)= c₁+c₂In1 =2

 

since In(1)=0, then c₁ =2.

y'(x)=-x^(-3/2)-(c₂/2)x^(-3/2)Inx+c₂x^(-3/2)

y'(1)=1

y'(1)=-1+c₂=1

c₂ = 2

 

the general solution is then

 

y(x) = 2x^(-1/2) +2x^(-1/2)Inx