Find the general solution of the differential equation: $y^{\left(4\right)}-2y^{\left(2\right)}+y=0$.

• A. $y\left(x\right)=C_1{e}^x+C_2{e}^{-x}+C_3 \cos \left(x\right)+C_4 \sin \left(x\right)$
• B. $y\left(x\right)=C_1{e}^{2x}+C_2{e}^{-2x}+C_3{e}^x+C_4{e}^{-x}$
• C. $y\left(x\right)=C_1{e}^x+C_2{e}^{-x}+C_{3}x{e}^x+C_{4}x{e}^{-x}$
• D. $y\left(x\right)=C_1\cos \left(2x\right)+C_2\sin \left(2x\right)+C_3\cos \left(x\right)+C_4\sin \left(x\right)$

Answer: (A)

Verified step by step guidance

1. Step 1: Recognize that the given differential equation is a linear homogeneous differential equation with constant coefficients. The general form is y^{(4)} - 2y'' + y = 0.
2. Step 2: Assume a solution of the form y = e^{rx}, where r is a constant. Substitute y = e^{rx} into the differential equation to obtain the characteristic equation: r^4 - 2r^2 + 1 = 0.
3. Step 3: Solve the characteristic equation r^4 - 2r^2 + 1 = 0. Notice that this is a quadratic equation in terms of r^2. Let u = r^2, so the equation becomes u^2 - 2u + 1 = 0. Factorize this as (u - 1)^2 = 0, giving u = 1. Since u = r^2, we have r^2 = 1, which implies r = ±1.
4. Step 4: The roots of the characteristic equation are r = 1, r = -1, each with multiplicity 2. For distinct roots, the solutions are e^{rx}. For repeated roots, we include terms with x multiplied by e^{rx}. Thus, the general solution is y(x) = C_1 e^{x} + C_2 e^{-x} + C_3 x e^{x} + C_4 x e^{-x}, where C_1, C_2, C_3, and C_4 are constants.
5. Step 5: Verify the solution by substituting y(x) = C_1 e^{x} + C_2 e^{-x} + C_3 x e^{x} + C_4 x e^{-x} back into the original differential equation. Simplify to confirm that it satisfies y^{(4)} - 2y'' + y = 0.