Consider the differential equation y" + 6y' + 25y - 0 with initial condition y(0) = 0. Then, the general solution of the IVP isa)e-3x (A cos 4x + B sin 4x)b)Be-3xsin 4xc)Ae-4x sin 3xd)e-4x (A cos 3x + B sin 3x)Correct answer is option 'B'. Can you explain this answer?

Question

Accepted Answer

we have (D² + 6D + 25)y = 0

So, y =e^-3x [A cos 4x + B sin 4x]
but y(0) = 0 implies A = 0
Hence, y = Be^-3x sin 4x

Answer

To solve the differential equation y'' - 4y' + 4y = 0, we can use the method of characteristic equation.

The characteristic equation is obtained by assuming a solution of the form y = e^(rx), where r is a constant. Substituting this into the differential equation, we get:

r^2e^(rx) - 4re^(rx) + 4e^(rx) = 0

Factoring out e^(rx), we have:

e^(rx)(r^2 - 4r + 4) = 0

Setting the expression inside the parentheses equal to zero, we obtain the characteristic equation:

r^2 - 4r + 4 = 0

This equation can be factored as:

(r - 2)^2 = 0

So, we have a repeated root r = 2.

Therefore, the general solution to the differential equation is given by:

y = C1e^(2x) + C2xe^(2x),

where C1 and C2 are constants.

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