Q12E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 11 and 12, use a substitution of the form to find a general solution to the given equation for x>c.

4(x+2)²y"+5y=0

Step-by-Step Solution

Answer

The general solution for the given equation is y= c₁ (x+2)^1/2 cos (ln x)+c₁ (x+2)^1/2 sin (ln x).

Define Cauchy-Euler equations:

In mathematics, a Cauchy problem is one in which the solution to a partial differential equation must satisfy specific constraints specified on a hypersurface in the domain. 

An initial value problem or a boundary value problem is an example of the Cauchy problem. 

The equation will be in the form of, ax²y"+bxy'+cy=0.

Find the general solution:

The given equation is,

4(x+2)²y"+5y=0

Let be the differential operator defined by the left-hand side of equation, that is

L [y](x)= 4(x+2)²y"+5y 

And set,

w(r,x) =(x+2)^r

Substituting the w(r,x) in place of y(x), you get

L [w](x)= 4(x+2)²((x+2)^r)"+5((x+2)^r)

=4(x+2)² [r(r-1)(r+2)^r-2]+5(x+2)^r

=(4r²-4r+5) (x+2)^r

Solving the indicial equation,

4r²-4r+5=0

r=(4±√(16-4× 4× 5))/8

=(4± 8i)/8

=1/2 ± i

The two roots are,

r₁=(x+2)^1/2 cos (ln x)  and  r₂= (x+2)^1/2 sin (ln x)

Thus there are two linearly independent solutions given by,

y₁ = c₁ (x+2)^1/2 cos (ln x)

y₂ = c₂ (x+2)^1/2 sin (ln x)

Therefore, the general solution for the equation will be,

y=c₁ (x+2)^1/2 cos (ln x) + c₂ (x+2)^1/2 sin (ln x)