Example 3. Convert the following initial value problem subject to the initial conditions y" - 3y" -6y + 5y = 0, y(0) = y' (0)=y" (0) = 1, to an equivalent Volterra integral equation. (92) (93)

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Example 3. Convert the following initial value problem
subject to the initial conditions
to an equivalent Volterra integral equation.
As indicated before, we first set
and
y" - 3y" -6y + 5y = 0,
y(0) = y' (0)=y" (0) = 1,
y(x) = u(x).
(94)
Integrating both sides of (94) from 0 to x and using the initial condition y"(0) = 1 we
find
and
Integrating (95) twice and using the proper initial conditions we find
f*f* u(t) dtdt
(x) = 1 +
u(x) = 4+x=
y (x) = 1 + x +
y'(x)=1
the equivalent Volterra integral equation.
y(x) = 1+x+
y(x) = 1+x+=x² +
- 322² + √²*²²* u(t)dtdtdt.
Transforming the double and triple integrals in (96) and (97) to single integrals by using
the formulas (71) and (72) we find
= 1+2+
fu(t)dt.
1
2
S (x- t)u(t)dt,
Substituting (94), (95), (98) and (99) into (92) we find
- 352x² + √² (3+6
x² +
(92)
1
2
S (r-t)²u(t)dt.
(93)
3+6(x-t) -
) - (x − 1)²) u(t)dt,
(95)
(96)
(97)
(98)
(99)
(100)
Transcribed Image Text:Example 3. Convert the following initial value problem subject to the initial conditions to an equivalent Volterra integral equation. As indicated before, we first set and y" - 3y" -6y + 5y = 0, y(0) = y' (0)=y" (0) = 1, y(x) = u(x). (94) Integrating both sides of (94) from 0 to x and using the initial condition y"(0) = 1 we find and Integrating (95) twice and using the proper initial conditions we find f*f* u(t) dtdt (x) = 1 + u(x) = 4+x= y (x) = 1 + x + y'(x)=1 the equivalent Volterra integral equation. y(x) = 1+x+ y(x) = 1+x+=x² + - 322² + √²*²²* u(t)dtdtdt. Transforming the double and triple integrals in (96) and (97) to single integrals by using the formulas (71) and (72) we find = 1+2+ fu(t)dt. 1 2 S (x- t)u(t)dt, Substituting (94), (95), (98) and (99) into (92) we find - 352x² + √² (3+6 x² + (92) 1 2 S (r-t)²u(t)dt. (93) 3+6(x-t) - ) - (x − 1)²) u(t)dt, (95) (96) (97) (98) (99) (100)
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