Сharacteristic equation:
9 k 2 − 4 = 0 9{k^2} - 4 = 0 9 k 2 − 4 = 0
k 2 = 4 9 {k^2} = \frac{4}{9} k 2 = 9 4
k = ± 2 3 k = \pm \frac{2}{3} k = ± 3 2
Then the general solution of the homogeneous equation is
y 0 = C 1 e 2 3 x + C 2 e − 2 3 x {y_0} = {C_1}{e^{\frac{2}{3}x}} + {C_2}{e^{ - \frac{2}{3}x}} y 0 = C 1 e 3 2 x + C 2 e − 3 2 x
We will seek a particular solution in the form
y ~ = A sin x + B cos x ⇒ y ~ ′ = A cos x − B sin x ⇒ y ~ ′ ′ = − A sin x − B cos x \widetilde y = A\sin x + B\cos x \Rightarrow {\widetilde y^\prime } = A\cos x - B\sin x \Rightarrow {\widetilde y^{\prime \prime }} = - A\sin x - B\cos x y = A sin x + B cos x ⇒ y ′ = A cos x − B sin x ⇒ y ′′ = − A sin x − B cos x
Substitute the obtained values into the original equation:
− 9 A sin x − 9 B cos x − 4 A sin x − 4 B cos x = sin x - 9A\sin x - 9B\cos x - 4A\sin x - 4B\cos x = \sin x − 9 A sin x − 9 B cos x − 4 A sin x − 4 B cos x = sin x
{ − 13 A = 1 − 13 B = 0 ⇒ A = − 1 13 , B = 0 \left\{ \begin{array}{l}
- 13A = 1\\
- 13B = 0
\end{array} \right. \Rightarrow A = - \frac{1}{{13}},\,B = 0 { − 13 A = 1 − 13 B = 0 ⇒ A = − 13 1 , B = 0
So,
y ~ = − 1 13 sin x \widetilde y = - \frac{1}{{13}}\sin x y = − 13 1 sin x
y = y 0 + y = C 1 e 2 3 x + C 2 e − 2 3 x − 1 13 sin x y = {y_0} + y = {C_1}{e^{\frac{2}{3}x}} + {C_2}{e^{ - \frac{2}{3}x}} - \frac{1}{{13}}\sin x y = y 0 + y = C 1 e 3 2 x + C 2 e − 3 2 x − 13 1 sin x
Answer: y = C 1 e 2 3 x + C 2 e − 2 3 x − 1 13 sin x y = {C_1}{e^{\frac{2}{3}x}} + {C_2}{e^{ - \frac{2}{3}x}} - \frac{1}{{13}}\sin x y = C 1 e 3 2 x + C 2 e − 3 2 x − 13 1 sin x
#339914 on Dec 2023