Differential Equation Exercise 21
We are asked to solve the following differential equation:
$$ y'' - y = 2x \sin(x) $$
This is a second-order nonhomogeneous differential equation.
The nonhomogeneous term is of the form Q(x)sin(βx), which makes it suitable for solving using the method of undetermined coefficients.
We'll solve it in three steps.
1] Solving the homogeneous equation
We begin by finding the general solution to the associated homogeneous equation:
$$ y'' - y = 2x \sin(x) $$
To do this, we introduce an auxiliary variable and form the characteristic equation:
$$ t^2 - 1 = 0 $$
This yields two distinct real roots: \( t_1 = -1 \) and \( t_2 = 1 \)
$$ t^2 = 1 $$
$$ \sqrt{t^2} = \sqrt{1} $$
$$ t = \begin{cases} t_1 = -1 \\ \\ t_2 = 1 \end{cases} $$
Since the roots are distinct, the general solution of the homogeneous equation is:
$$ y_o = c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x} $$
where λ1 = -1 and λ2 = 1
$$ y_o = c_1 e^{-x} + c_2 e^{x} $$
This gives us the homogeneous solution.
Next, we compute a particular solution to the nonhomogeneous equation.
2] Finding a particular solution
To find a particular solution, we apply the method of undetermined coefficients.
The right-hand side is of the form Q(x)sin(βx), with Q(x) = 2x and β = 1.
We first check whether \( i\beta \) is a root of the characteristic equation:
$$ t^2 - 1 = 0 $$
$$ (i\beta)^2 - 1 = 0 $$
$$ (i \cdot 1)^2 - 1 = 0 $$
$$ i^2 - 1 = -1 - 1 = -2 \neq 0 $$
Since \( i\beta \) is not a root, the particular solution has the form:
$$ y_p = R(x) \cos(\beta x) + S(x) \sin(\beta x) $$
With β = 1, this becomes:
$$ y_p = R(x) \cos(x) + S(x) \sin(x) $$
Because Q(x) = 2x is a polynomial of degree 1, we assume R(x) = A + Bx and S(x) = C + Dx, both polynomials of degree 1:
$$ y_p = (A + Bx) \cos(x) + (C + Dx) \sin(x) $$
The constants A, B, C, and D are to be determined.
We now compute the first derivative of \( y_p \):
$$ y'_p = D_x [ (A + Bx) \cos(x) + (C + Dx) \sin(x) ] $$
Applying the product rule:
$$ y'_p = B \cos(x) - (A + Bx) \sin(x) + D \sin(x) + (C + Dx) \cos(x) $$
Now compute the second derivative:
$$ y''_p = D_x [ B \cos(x) - (A + Bx) \sin(x) + D \sin(x) + (C + Dx) \cos(x) ] $$
After differentiating:
$$ y''_p = -2B \sin(x) + 2D \cos(x) - (A + Bx) \cos(x) - (C + Dx) \sin(x) $$
Now we substitute \( y_p \) and \( y''_p \) into the original equation:
$$ y'' - y = 2x \sin(x) $$
Substituting gives:
$$ y''_p - y_p = 2x \sin(x) $$
Plugging in the expressions for \( y''_p \) and \( y_p \):
$$ [-2B \sin(x) + 2D \cos(x) - (A + Bx) \cos(x) - (C + Dx) \sin(x)] - [(A + Bx) \cos(x) + (C + Dx) \sin(x)] = 2x \sin(x) $$
Grouping like terms:
$$ \sin(x)[-2B - 2C - 2Dx] + \cos(x)[2D - 2A - 2Bx] = 2x \sin(x) $$
Now we match coefficients of \( \sin(x) \) and \( \cos(x) \) on both sides:
$$ \begin{cases} -2B - 2C - 2Dx = 2x \\ \\ 2D - 2A - 2Bx = 0 \end{cases} $$
Note: On the left-hand side, the coefficient of \( \sin(x) \) is \( -2B - 2C - 2Dx \); on the right-hand side, it's \( 2x \). For \( \cos(x) \), the left-hand coefficient is \( 2D - 2A - 2Bx \), and the right-hand coefficient is 0.
We equate the coefficients of like powers of \( x \):
$$ \begin{cases} -2D = 2 \\ -2B - 2C = 0 \\ -2B = 0 \\ 2D - 2A = 0 \end{cases} $$
Note: In the first equation, the coefficient of \( x \) in the first row gives \( -2D = 2 \). From the constant terms, we get the remaining equations.
Solving this system:
From \( -2D = 2 \) we get \( D = -1 \)
From \( -2B = 0 \), we find \( B = 0 \)
Plugging into \( -2B - 2C = 0 \) gives \( C = 0 \)
From \( 2D - 2A = 0 \) we find \( A = -1 \)
Now substitute these constants into the particular solution:
$$ y_p = (A + Bx) \cos(x) + (C + Dx) \sin(x) $$
$$ y_p = (-1) \cos(x) + (-x) \sin(x) $$
Therefore, the particular solution is:
$$ y_p = - \cos(x) - x \sin(x) $$
3] General solution
The general solution is the sum of the homogeneous and particular solutions:
$$ y = y_o + y_p $$
We know:
$$ y_o = c_1 e^{-x} + c_2 e^{x} $$
and
$$ y_p = - \cos(x) - x \sin(x) $$
Thus, the general solution is:
$$ y = c_1 e^{-x} + c_2 e^{x} - \cos(x) - x \sin(x) $$
where \( c_1 \) and \( c_2 \) are arbitrary real constants.
And that completes the solution.
