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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)=?

Respuesta :

Answer:

Step-by-step explanation:

Check part

[tex]y= C_1y_1 + C_2y_2 = C_1cos(7x)+C_2sin(7x)[/tex]

[tex]y'= -7 C_1sin(7x)+7C_2cos(7x)[/tex]

[tex]y"= -49 C_1cos(7x) - 49 C_2sin(7x)[/tex]

Now, replace to the original one.

[tex]y"+49 y = -49C_1cos(7x)-49 C_2 sin(7x) + 49 C_1cos(7x) +49 C_2sin(7x) = 0\\[/tex]

Done!!

Particular solution

[tex]y(0) = C_1cos(0) + C_2 sin(0) = C_1= 10[/tex]

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

[tex]y(0)'= -7 C_1sin(0) + 7 C_2 cos (0) = 7 C_2= -4[/tex]

[tex]C_2 = -4/7[/tex]

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

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