Question

A homogeneous second-order linear differential equation, two functions y1 and y2 , and a pair of initial conditions are given below. First verify that y1 and y2 are solutions of the differential equation. Then find a particular solution of the form
y = c1y1 + c2y2 that satisfies the given initial conditions.
y'' + 49y = 0; y1 = cos(7x) y2 = sin(7x); y(0) = 10 y(0)=-4
y(x)=?

Answer 1

Answer:

Step-by-step explanation:

Check part

$y= C_1y_1 + C_2y_2 = C_1cos(7x)+C_2sin(7x)$

$y'= -7 C_1sin(7x)+7C_2cos(7x)$

$y"= -49 C_1cos(7x) - 49 C_2sin(7x)$

Now, replace to the original one.

$y"+49 y = -49C_1cos(7x)-49 C_2 sin(7x) + 49 C_1cos(7x) +49 C_2sin(7x) = 0$

Done!!

Particular solution

$y(0) = C_1cos(0) + C_2 sin(0) = C_1= 10$

I believe that y'(0) = 4, not y(0) anymore. Since y(0) CANNOT have two different solution.

$y(0)'= -7 C_1sin(0) + 7 C_2 cos (0) = 7 C_2= -4$

$C_2 = -4/7$

The last step is to put C1, C2 into your solution. You finish it.