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Find the general solution of the second order, linear differential equationy′′ + 3y′ + 2y = sin(x), y = y(x)

Question

Find the general solution of the second order, linear differential equation

y+3y+2y=sin(x),y=y(x) y'' + 3y' + 2y = \sin(x), \quad y = y(x)

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Solution

Sure, let's solve this step by step.

Step 1: Solve the homogeneous equation The homogeneous equation is y'' + 3y' + 2y = 0. This is a second order linear differential equation with constant coefficients. The characteristic equation is r^2 + 3r + 2 = 0. Solving this quadratic equation gives r = -1, -2. Therefore, the general solution of the homogeneous equation is y_h(x) = C1e^(-x) + C2e^(-2x).

Step 2: Find a particular solution We use the method of undetermined coefficients to find a particular solution of the non-homogeneous equation. We guess a solution of the form y_p(x) = Acos(x) + Bsin(x). Differentiating this and substituting into the differential equation gives us two equations for A and B. Solving these equations gives us the coefficients A and B.

Step 3: Write down the general solution The general solution of the non-homogeneous equation is the sum of the general solution of the homogeneous equation and the particular solution. Therefore, the general solution is y(x) = C1e^(-x) + C2e^(-2x) + Acos(x) + Bsin(x).

Note: The actual values of A and B depend on the specific form of the non-homogeneous term and can be found by solving the system of equations obtained in step 2.

This problem has been solved

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