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Expert-verified Found in: Page 453 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # 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.

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).

See the step by step solution

## 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) = xr

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

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

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

=(r (r-1) (r-2)) xr-3+6 (r (r-1)) xr+29rxr -29 xr

=(r (r-1) (r-2)) xr-3+6 (r (r-1)) xr+29 (r-1) xr

=(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) ### Want to see more solutions like these? 