Question

Y ′′+6y +9y=17cos(2x)−7sin(2x), med y(0)=0,y' (0)=6.

Answer 1

The solution to the given differential equation with the specified initial conditions is
`\( y(x) = 3\cos(2x) + 2\sin(2x) \).`

Step-by-step explanation

To solve the differential equation
`\( y'' + 6y + 9y = 17\cos(2x) - 7\sin(2x) \),`we first find the complementary solution to the homogeneous equation \y'' + 6y + 9y = 0 . The characteristic equation is
`\( r^2 + 6r + 9 = 0 \),`which factors to
`\( (r + 3)^2 = 0 \).` This gives a repeated root r = -3 . The complementary solution is then
`\( y_c(x) = (C_1 + C_2x)e^(-3x) \).`

Next, we find the particular solution for the non-homogeneous part
`\( 17\cos(2x) - 7\sin(2x) \). Assuming \( y_p(x) = A\cos(2x) + B\sin(2x) \)`, we substitute it into the differential equation and solve for the coefficients A and B. The particular solution is
`\( y_p(x) = 3\cos(2x) - 2\sin(2x) \)`

Combining the complementary and particular solutions, we get the general solution
`\( y(x) = (C_1 + C_2x)e^(-3x) + 3\cos(2x) - 2\sin(2x) \).` Applying the initial conditions
`\( y(0) = 0 \) and \( y'(0) = 6 \), we find \( C_1 = 0 \) and \( C_2 = 6 \).` Therefore, the final solution is
`\( y(x) = 3\cos(2x) - 2\sin(2x) \).`