The solution of an equation u_(yy)+16 u=0 is (a) c_(1)(y)cos 4y+c_(2)(y)sin 4y (b) c_(1)(x)cos 4y+c_(2)(x)sin 4y (c) c_(1)(x)cos 4x+c_(2)(x)sin 4x (d) c_(1)(x)cos 4y+c_(2)(x)sin 4x

Question

The solution of an equation  u_(yy)+16 u=0 is (a)  c_(1)(y)cos 4y+c_(2)(y)sin 4y (b)  c_(1)(x)cos 4y+c_(2)(x)sin 4y (c)  c_(1)(x)cos 4x+c_(2)(x)sin 4x (d)  c_(1)(x)cos 4y+c_(2)(x)sin 4x

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Given Differential Equation: We are given the second-order linear homogeneous differential equation with constant coefficients: u_(yy) + 16u = 0 To solve this, we look for solutions of the form u(y) = e^(λy), where λ is a constant to be determined. Substituting u(y) into the differential equation gives us: (λ^2)e^(λy) + 16e^(λy) = 0 Dividing through by e^(λy), which is never zero, we get: λ^2 + 16 = 0Solution Form: We now solve the characteristic equation for λ: λ^2 + 16 = 0 λ^2 = -16 λ = ±4i where i is the imaginary unit.Characteristic Equation: The general solution to the differential equation with complex roots λ = α ± βi is given by: u(y) = e^(αy)(c_1 cos(βy) + c_2 sin(βy)) Since α = 0 and β = 4 in our case, the solution simplifies to: u(y) = c_1 cos(4y) + c_2 sin(4y)General Solution: The solution u(y) is a function of y, and the coefficients c_1 and c_2 are arbitrary constants that do not depend on y. Therefore, the correct form of the solution is: u(y) = c_1(y) cos(4y) + c_2(y) sin(4y) However, since c_1 and c_2 are constants, not functions of y, the correct notation should be: u(y) = c_1 cos(4y) + c_2 sin(4y)Correct Form: Looking at the options given, we see that option (a) incorrectly suggests c_1 and c_2 are functions of y, and options (b) and (d) incorrectly introduce the variable x, which is not present in the original differential equation. The correct option is (c), which correctly identifies c_1 and c_2 as constants and maintains y as the variable.

The solution of an equation $u_{y y}+16 u=0$ is $\newline$ (a) $c_{1}(y) \cos 4 y+c_{2}(y) \sin 4 y$ $\newline$ (b) $c_{1}(x) \cos 4 y+c_{2}(x) \sin 4 y$ $\newline$ (c) $c_{1}(x) \cos 4 x+c_{2}(x) \sin 4 x$ $\newline$ (d) $c_{1}(x) \cos 4 y+c_{2}(x) \sin 4 x$

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