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Q17E

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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)