Answer on Question #60151 – Math – Differential Equations
Question
Solve the following differential equation y ′ ′ + y = cos 2 x y'' + y = \cos^2 x y ′′ + y = cos 2 x
Solution
The homogeneous second order linear differential equation is
y ′ ′ + y = 0. y'' + y = 0. y ′′ + y = 0.
It is a differential equation with constant coefficients.
The characteristic polynomial of the differential equation (1) is
r 2 + 1 = 0. r^2 + 1 = 0. r 2 + 1 = 0.
There are two complex conjugate roots:
r 1 = i and r 2 = − i . r_1 = i \text{ and } r_2 = -i. r 1 = i and r 2 = − i .
Hence the general solution of the homogeneous equation (1) is
y h = C 1 cos x + C 2 sin x , y_h = C_1 \cos x + C_2 \sin x, y h = C 1 cos x + C 2 sin x ,
where C 1 , C 2 C_1, C_2 C 1 , C 2
Using power-reducing/half angle formulas cos 2 x = 1 + cos ( 2 x ) 2 \cos^2 x = \frac{1 + \cos(2x)}{2} cos 2 x = 2 1 + c o s ( 2 x ) y ′ ′ + y = cos 2 x y'' + y = \cos^2 x y ′′ + y = cos 2 x
y ′ ′ + y = 1 2 + 1 2 cos ( 2 x ) y'' + y = \frac{1}{2} + \frac{1}{2} \cos(2x) y ′′ + y = 2 1 + 2 1 cos ( 2 x )
A particular solution of (2) will be sought in the form
y p = A + B cos ( 2 x ) + C sin ( 2 x ) , y_p = A + B \cos(2x) + C \sin(2x), y p = A + B cos ( 2 x ) + C sin ( 2 x ) ,
where A , B , C A, B, C A , B , C
Differentiating (3) with respect to x x x
y p ′ = − 2 B sin ( 2 x ) + 2 C cos ( 2 x ) y p ′ ′ = − 4 B cos ( 2 x ) − 4 C sin ( 2 x ) \begin{aligned}
y_p' &= -2B \sin(2x) + 2C \cos(2x) \\
y_p'' &= -4B \cos(2x) - 4C \sin(2x)
\end{aligned} y p ′ y p ′′ = − 2 B sin ( 2 x ) + 2 C cos ( 2 x ) = − 4 B cos ( 2 x ) − 4 C sin ( 2 x )
Substituting these derivatives in equation (2) get
− 4 B cos ( 2 x ) − 4 C sin ( 2 x ) + A + B cos ( 2 x ) + C sin ( 2 x ) = 1 2 + 1 2 cos ( 2 x ) -4B \cos(2x) - 4C \sin(2x) + A + B \cos(2x) + C \sin(2x) = \frac{1}{2} + \frac{1}{2} \cos(2x) − 4 B cos ( 2 x ) − 4 C sin ( 2 x ) + A + B cos ( 2 x ) + C sin ( 2 x ) = 2 1 + 2 1 cos ( 2 x )
Equating the corresponding coefficients obtain
{ A = 1 2 − 3 B = 1 2 − 3 C = 0 ⇒ { A = 1 2 B = − 1 6 C = 0 \left\{
\begin{array}{l}
A = \frac{1}{2} \\
-3B = \frac{1}{2} \\
-3C = 0
\end{array}
\right.
\Rightarrow
\left\{
\begin{array}{l}
A = \frac{1}{2} \\
B = -\frac{1}{6} \\
C = 0
\end{array}
\right. ⎩ ⎨ ⎧ A = 2 1 − 3 B = 2 1 − 3 C = 0 ⇒ ⎩ ⎨ ⎧ A = 2 1 B = − 6 1 C = 0
Thus,
y p = 1 2 − 1 6 cos ( 2 x ) . y_p = \frac{1}{2} - \frac{1}{6} \cos(2x). y p = 2 1 − 6 1 cos ( 2 x ) .
The general solution of the equation (2) is the sum of a particular solution y p y_p y p y h y_h y h
Therefore,
y = C 1 cos x + C 2 sin x + 1 2 − 1 6 cos ( 2 x ) , y = C_1 \cos x + C_2 \sin x + \frac{1}{2} - \frac{1}{6} \cos(2x), y = C 1 cos x + C 2 sin x + 2 1 − 6 1 cos ( 2 x ) ,
where C 1 , C 2 C_1, C_2 C 1 , C 2
**Answer**: y = C 1 cos x + C 2 sin x + 1 2 − 1 6 cos ( 2 x ) y = C_1 \cos x + C_2 \sin x + \frac{1}{2} - \frac{1}{6} \cos(2x) y = C 1 cos x + C 2 sin x + 2 1 − 6 1 cos ( 2 x )
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