Question

3. Show that y=(1/4) sin 4x is a unique solution of the initial value problem
y" + 16y=0 with y(0) = 0 and y'(0) = 1.
(1) =3 and y'(1) = 1​

Answer 1

Answer:

Step-by-step explanation:

where p and q are continuous functions on some interval I. A second order, linear

differential equation has an analogous form.

DEFINITION 1. A second order linear differential equation is an equation which can

be written in the form

y00 + p(x)y0 + q(x)y = f(x) (1)

where p, q, and f are continuous functions on some interval I.

The functions p and q are called the coefficients of the equation; the function f on

the right-hand side is called the forcing function or the nonhomogeneous term . The term

“forcing function” comes from applications of second-order linear equations; the description

“nonhomogeneous” is given below.

A second order equation which is not linear is said to be nonlinear .

Examples

(a) y00 − 5y0 + 6y = 3 cos 2x. Here p(x) = −5, q(x)=6, f(x) = 3 cos 2x are

continuous functions on (−∞, ∞).

(b) x2 y00 − 2x y0 + 2y = 0. This equation is linear because it can be written in the form

(1) as

y00 − 2

x

y0 +

2

x2 y = 0

where p(x)=2/x, q(x)=2/x2, f(x) = 0 are continuous on any interval that does

not contain x = 0. For example, we could take I = (0, ∞).

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(c) y00 +xy2y0 −y3 = exy is a nonlinear equation; this equation cannot be written in the

form (1).

Remarks on “Linear.” Intuitively, a second order differential equation is linear if y00

appears in the equation with exponent 1 only, and if either or both of y and y0 appear

in the equation, then they do so with exponent 1 only. Also, there are no so-called “crossproduct” terms, y y0

,yy00, y0 y00. In this sense, it is easy to see that the equations in (a) and

(b) are linear, and the equation in (c) is nonlinear.

Set L[y] = y00 + p(x)y0 + q(x)y. If we view L as an “operator” that transforms a twice

differentiable function y = y(x) into the continuous function

L[y(x)] = y00(x) + p(x)y0

(x) + q(x)y(x),

then, for any two twice differentiable functions y1(x) and y2(x),

L[y1(x) + y2(x)] = [y1(x) + y2(x)]00 + p(x)[y1(x) + y2(x)]0 + q(x)[y1(x) + y2(x)]

= y00

1 (x) + y00

2 (x) + p(x)[y0

1(x) + y0

2(x)] + q(x)[y1(x) + y2(x)]

= y00

1 (x) + p(x)y0

1(x) + q(x)y1(x) + y00

2 (x) + p(x)y0

2(x) + q(x) + y2(x)

(x) =