Use the method of "undetermined coefficients" to solve the 2nd ODE #y''+4y'+4y=e^(-2x)sin2x# ?
# y(x) = Axe^(-2x)+Be^(-2x) -1/4e^(-2x)sin2x #
We have:
Complementary Role
The equation homogeneous linked to [A] is
Additionally, the related auxiliary equation is:
Consequently, the homogeneous equation's solution is:
Specific Resolution
For this specific equation [A], the following is a likely solution:
And:
When we solve concurrently, we have:
Thus, we arrive at the specific solution:
Overall Resolution
which ultimately results in the GS of [A}
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To solve the second-order ordinary differential equation (ODE) (y'' + 4y' + 4y = e^{-2x}\sin(2x)) using the method of undetermined coefficients, follow these steps:
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Find the complementary solution by solving the homogeneous equation (y'' + 4y' + 4y = 0). The characteristic equation is (r^2 + 4r + 4 = 0), which has a repeated root at (r = -2). So, the complementary solution is (y_c(x) = (C_1 + C_2x)e^{-2x}).
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Next, consider the particular solution (y_p(x)) for the non-homogeneous part (e^{-2x}\sin(2x)). Since (e^{-2x}\sin(2x)) is a product of an exponential function and a trigonometric function, we assume (y_p(x) = Ae^{-2x}\sin(2x) + Be^{-2x}\cos(2x)), where (A) and (B) are undetermined coefficients.
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Differentiate (y_p(x)) twice to find (y_p'') and (y_p').
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Substitute (y_p), (y_p'), and (y_p'') into the original ODE (y'' + 4y' + 4y = e^{-2x}\sin(2x)).
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Equate the coefficients of like terms on both sides of the equation.
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Solve the resulting system of equations to find the values of (A) and (B).
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Combine the complementary solution (y_c(x)) and the particular solution (y_p(x)) to get the general solution (y(x) = y_c(x) + y_p(x)).
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Finally, if initial conditions are given, use them to determine the values of the constants (C_1) and (C_2).
This completes the solution process.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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