Q17E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 15-17,solve the given initial value problem x³y"'+6x²y"+29xy'-29y=0  y(1)=1 and y'(1)= -3 and y"(1)=19.

Step-by-Step Solution

Answer

The solution of the given initial value problem is y= 31/17 x + 3/17 x^-2 cos (5 lnx) -76/85 x^-2 sin (5 lnx).

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,

x³y"'+6x²y"+29xy'-29y=0

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

L [x] (x) =x³y"'+6x²y"+29xy'-29y

And let.

w (r,x) = x^r

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

L [w] (x) = x³ (x^r)"'+6x²(x^r) "+29x(x^r)'-29(x^r) 

=x³ (r(r-1) (r-2)) x^r-3 + 6x² (r (r-1)) x^r-2 +29 x(r) x^r-1 -29 x^r

=(r (r-1) (r-2)) x^r-3+6 (r (r-1)) x^r+29rx^r -29 x^r 

=(r (r-1) (r-2)) x^r-3+6 (r (r-1)) x^r+29 (r-1) x^r

=(r-1) (r²+4r+29)

√

Solving the quadratic part of the above indicial equation,

r²+4r+29=0

r= (-4± √(16-116)) /2

= (-4± 10i)/2

= -2± 5i

Thus the indicial equation has one real root and two complex conjugates,

r₁ =1

r₂= -2+5i

r₃= -2-5i

For complex roots:

x^-2+5i = x^-2 cos (5 lnx)+ix^-2 sin(5 lnx)

We can write two linearly independent solutions as,

y=x^-2 cos (5 lnx)  and  y=x^-2 sin (5 lnx)

Thus the the three linearly independent real-valued solutions are,

y₁=x ,  y₂= x^-2 cos(5 lnx) and y₃=x^-2 sin (5 lnx)

The general solution for the equation will be,

y = c₁x+c₂ x^-2 cos(5 lnx) +c₂ x^-2 sin (5 lnx)

Determine the initial value:

For the given initial conditions:

y(1) =2, y'(1) = -3  and y"(1)=19

y(x) = c₁x + c₂x^-2 cos(5 lnx) +c₂ x^-2 sin (5 lnx)

y(1) =  c₁+ c₂(1)^-2 cos(5 ln (1)) +c₂ (1)^-2 sin (5 ln (1))

y(1) = c₁ + c₂ x^-2 cos(0)+c₂ x^-2 cos(0)+c₂ x^-2 sin(0)

y(1)=c₁+c₂

c₁+c₂=2                                                                                                               ..... (1)

y'(x) = c₁+x^-3 cos(5 lnx) (-2c₂+5c₃) +x^-3 sin(5 lnx) (-5c₂-2c₃)

y'(1) = c₁+(1)^-3 cos(5 ln (1)) (-2c₂+5c₃) +x^-3 sin(5 ln (1)) (-5c₂-2c₃)

y'(1) = c₁-2c₂+5c₃

c₁-2c₂+5c₃= -3                                                                                                    ..... (2)

y"(x) = x^-4 cos(5 lnx) (-19c₂-25c₃) +x^-3 sin(5 lnx) (5c₂+31c₃)

y"(1) = (1)^-4 cos(5 ln (1)) (-19c₂-25c₃) +(1)^-3 sin(5 lnx) (5c₂+31c₃)

y"(1) = -19c₂ -25c₃

 -19c₂ -25c₃ = 19                                                                                                 ..... (3)

Solving equations (1), (2) and (3) using matrix reduction method,

Step 1: R₃  → R₃ +5R₂

Step 2: R₃  → R₃ +29 R₁

Re-writing the equations

c₁+c₂=2

c₁-2c₂+5c₃ = -3

34c₁ = 62

From above, we get the constants c₁, c₂, and c₃ as,

c₁ = 31/17 ; c₂=3/17 ; c₃ = -76/85

The solution for the equation will be,

y= 31/17 x + 3/17 x^-2 cos (5 lnx) -76/85 x^-2 sin (5 lnx)