Question

(D^2+9) y=cos3x find the general solution

Answer 1

Answer:

On the right hand side of the differential equation one “^” sign is there. So, I've assumed the RHS of the equation as cos^3 x (that is, cos cube x and not cos 3x).

Answer 2

General solution is y = c1 sin(3x) + c2 cos(3x) + (1/9)sin(3x)

Given:

(D^2 + 9)y = cos(3x)

To find:

General solution.

Solution:

Given in the question,

(D^2 + 9)y = cos(3x)

The complementary function of the given equation,

(D^2 + 9)y = 0

Now,

Characteristic equation ,

r^2 + 9 = 0

⇒r^2= 9

⇒r = ±3i

Now,

y = c1 sin(3x) + c2 cos(3x)

Here,

c1 and c2 = constants.

Let the equation be,

y = A cos(3x) + B sin(3x)

Here,

A and B = Constants.

Now,

If we differentiating y with respect to x,

y' = -3A sin(3x) + 3B cos(3x)

Again,

y'' = -9A cos(3x) - 9B sin(3x)

Now after putting the value of y, y', and y'' into the equation,

(-9A cos(3x) - 9B sin(3x) + 9A cos(3x) + 9B sin(3x)) + 9(A cos(3x) + B sin(3x)) = cos(3x)

⇒9(A cos(3x) + B sin(3x)) = cos(3x)

So,

A = 0

B = 1/9.

The particular integral ,

y = (1/9)sin(3x)

General Solution,

y = c1 sin(3x) + c2 cos(3x) + (1/9)sin(3x)

Here c1 and c2 are constants.

General solution is y = c1 sin(3x) + c2 cos(3x) + (1/9)sin(3x)

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