Question1
1) 1 50 q ′ ′ + q ′ + 8 q = 50 cos 30 t \frac{1}{{50}}q'' + q' + 8q = 50\cos 30t 50 1 q ′′ + q ′ + 8 q = 50 cos 30 t
q ′ ′ + 50 q ′ + 400 q = 2500 cos 30 t q'' + 50q' + 400q = 2500\cos 30t q ′′ + 50 q ′ + 400 q = 2500 cos 30 t
characteristic equation:
k 2 + 50 k + 400 = 0 {k^2} + 50k + 400 = 0 k 2 + 50 k + 400 = 0
D = 2500 − 1600 = 900 D = 2500 - 1600 = 900 D = 2500 − 1600 = 900
k 1 = − 50 − 30 2 = − 40 {k_1} = \frac{{ - 50 - 30}}{2} = - 40 k 1 = 2 − 50 − 30 = − 40
k 2 = − 50 + 30 2 = − 10 {k_2} = \frac{{ - 50 + 30}}{2} = - 10 k 2 = 2 − 50 + 30 = − 10
Then the general solution of the homogeneous equation is
q 0 = C 1 e − 40 t + C 2 e − 10 t {q_0} = {C_1}{e^{ - 40t}} + {C_2}{e^{ - 10t}} q 0 = C 1 e − 40 t + C 2 e − 10 t
We will seek a particular solution in the form
q ~ = A cos 30 t + B sin 30 t ⇒ q ~ ′ = − 30 A sin 30 t + 30 B cos 30 t ⇒ q ~ ′ ′ = = − 900 A cos 30 t − 900 B sin 30 t \widetilde q = A\cos 30t + B\sin 30t \Rightarrow {\widetilde q^\prime } = - 30A\sin 30t + 30B\cos 30t \Rightarrow {\widetilde q^{\prime \prime }} = \\=- 900A\cos 30t - 900B\sin 30t q = A cos 30 t + B sin 30 t ⇒ q ′ = − 30 A sin 30 t + 30 B cos 30 t ⇒ q ′′ = = − 900 A cos 30 t − 900 B sin 30 t
Substitute the obtained values into the original equation:
− 900 A cos 30 t − 900 B sin 30 t − 1500 A sin 30 t + 1500 B cos 30 t + 400 A cos 30 t + 400 B sin 30 t = 2500 cos 30 t - 900A\cos 30t - 900B\sin 30t - 1500A\sin 30t + 1500B\cos 30t + 400A\cos 30t + 400B\sin 30t = 2500\cos 30t − 900 A cos 30 t − 900 B sin 30 t − 1500 A sin 30 t + 1500 B cos 30 t + 400 A cos 30 t + 400 B sin 30 t = 2500 cos 30 t
( − 900 A + 1500 B + 400 A ) cos 30 t + ( − 900 B − 1500 A + 400 B ) sin 30 t = 2500 cos 30 t ( - 900A + 1500B + 400A)\cos 30t + ( - 900B - 1500A + 400B)\sin 30t = 2500\cos 30t ( − 900 A + 1500 B + 400 A ) cos 30 t + ( − 900 B − 1500 A + 400 B ) sin 30 t = 2500 cos 30 t
{ − 500 A + 1500 B = 2500 − 1500 A − 500 B = 0 ⇒ \left\{ \begin{array}{l}
- 500A + 1500B = 2500\\
- 1500A - 500B = 0
\end{array} \right. \Rightarrow { − 500 A + 1500 B = 2500 − 1500 A − 500 B = 0 ⇒
⇒ { A = − 1 2 B = 3 2 \Rightarrow \left\{ {\begin{matrix}
{A = - \frac{1}{2}}\\
{B = \frac{3}{2}}
\end{matrix}} \right. ⇒ { A = − 2 1 B = 2 3
So,
q ~ = − 1 2 cos 30 t + 3 2 sin 30 t \widetilde q = - \frac{1}{2}\cos 30t + \frac{3}{2}\sin 30t q = − 2 1 cos 30 t + 2 3 sin 30 t
q ( t ) = q 0 + q ~ = C 1 e − 40 t + C 2 e − 10 t − 1 2 cos 30 t + 3 2 sin 30 t q(t) = {q_0} + \widetilde q = {C_1}{e^{ - 40t}} + {C_2}{e^{ - 10t}} - \frac{1}{2}\cos 30t + \frac{3}{2}\sin 30t q ( t ) = q 0 + q = C 1 e − 40 t + C 2 e − 10 t − 2 1 cos 30 t + 2 3 sin 30 t
Answer: q ( t ) = C 1 e − 40 t + C 2 e − 10 t − 1 2 cos 30 t + 3 2 sin 30 t q(t) = {C_1}{e^{ - 40t}} + {C_2}{e^{ - 10t}} - \frac{1}{2}\cos 30t + \frac{3}{2}\sin 30t q ( t ) = C 1 e − 40 t + C 2 e − 10 t − 2 1 cos 30 t + 2 3 sin 30 t
2)
I ( t ) = d q d t = ( C 1 e − 40 t + C 2 e − 10 t − 1 2 cos 30 t + 3 2 sin 30 t ) ′ = = − 40 C 1 e − 40 t − 10 C 2 e − 10 t + 15 sin 30 t + 45 cos 30 t I(t) = \frac{{dq}}{{dt}} = {\left( {{C_1}{e^{ - 40t}} + {C_2}{e^{ - 10t}} - \frac{1}{2}\cos 30t + \frac{3}{2}\sin 30t} \right)^\prime } =\\= - 40{C_1}{e^{ - 40t}} - 10{C_2}{e^{ - 10t}} + 15\sin 30t + 45\cos 30t I ( t ) = d t d q = ( C 1 e − 40 t + C 2 e − 10 t − 2 1 cos 30 t + 2 3 sin 30 t ) ′ = = − 40 C 1 e − 40 t − 10 C 2 e − 10 t + 15 sin 30 t + 45 cos 30 t
Answer: I ( t ) = − 40 C 1 e − 40 t − 10 C 2 e − 10 t + 15 sin 30 t + 45 cos 30 t I(t) = - 40{C_1}{e^{ - 40t}} - 10{C_2}{e^{ - 10t}} + 15\sin 30t + 45\cos 30t I ( t ) = − 40 C 1 e − 40 t − 10 C 2 e − 10 t + 15 sin 30 t + 45 cos 30 t
3) The steady state solution is q ~ = − 1 2 cos 30 t + 3 2 sin 30 t = 5 2 sin ( 30 t − a r c t a n 1 3 ) \widetilde q = - \frac{1}{2}\cos 30t + \frac{3}{2}\sin 30t = \sqrt {\frac{5}{2}} \sin \left( {30t - arctan\frac{1}{3}} \right) q = − 2 1 cos 30 t + 2 3 sin 30 t = 2 5 sin ( 30 t − a rc t an 3 1 )
So, the amplitude is A = 5 2 A = \sqrt {\frac{5}{2}} A = 2 5 f = 30 2 π f =\frac{ 30}{2\pi} f = 2 π 30
Answer: A = 5 2 A = \sqrt {\frac{5}{2}} A = 2 5 f = 30 2 π f = \frac{30}{2 \pi} f = 2 π 30
Question 2
y ′ ′ + 2 y ′ + 5 y = 34 sin x cos x y'' + 2y' + 5y = 34\sin x\cos x y ′′ + 2 y ′ + 5 y = 34 sin x cos x
y ′ ′ + 2 y ′ + 5 y = 17 sin 2 x y'' + 2y' + 5y = 17\sin 2x y ′′ + 2 y ′ + 5 y = 17 sin 2 x
characteristic equation:
k 2 + 2 k + 5 = 0 {k^2} + 2k + 5 = 0 k 2 + 2 k + 5 = 0
D = 4 − 20 = − 16 D = 4 - 20 = - 16 D = 4 − 20 = − 16
k 1 = − 2 − 4 i 2 = − 1 − 2 i {k_1} = \frac{{ - 2 - 4i}}{2} = - 1 - 2i k 1 = 2 − 2 − 4 i = − 1 − 2 i
k 2 = − 2 + 4 i 2 = − 1 + 2 i {k_2} = \frac{{ - 2 + 4i}}{2} = - 1 + 2i k 2 = 2 − 2 + 4 i = − 1 + 2 i
Then the general solution of the homogeneous equation is
y 0 = e − x ( C 1 cos 2 x + C 2 sin 2 x ) {y_0} = {e^{ - x}}\left( {{C_1}\cos 2x + {C_2}\sin 2x} \right) y 0 = e − x ( C 1 cos 2 x + C 2 sin 2 x )
We will seek a particular solution in the form
y ~ = A sin 2 x + B cos 2 x ⇒ y ~ ′ = 2 A cos 2 x − 2 B sin 2 x ⇒ y ~ ′ ′ = − 4 A sin 2 x − 4 B cos 2 x \widetilde y = A\sin 2x + B\cos 2x \Rightarrow {\widetilde y^\prime } = 2A\cos 2x - 2B\sin 2x \Rightarrow {\widetilde y^{\prime \prime }} = - 4A\sin 2x - 4B\cos 2x y = A sin 2 x + B cos 2 x ⇒ y ′ = 2 A cos 2 x − 2 B sin 2 x ⇒ y ′′ = − 4 A sin 2 x − 4 B cos 2 x
Substitute the obtained values into the original equation:
− 4 A sin 2 x − 4 B cos 2 x + 4 A cos 2 x − 4 B sin 2 x + 5 A sin 2 x + 5 B cos 2 x = 17 sin 2 x - 4A\sin 2x - 4B\cos 2x + 4A\cos 2x - 4B\sin 2x + 5A\sin 2x + 5B\cos 2x = 17\sin 2x − 4 A sin 2 x − 4 B cos 2 x + 4 A cos 2 x − 4 B sin 2 x + 5 A sin 2 x + 5 B cos 2 x = 17 sin 2 x
( A − 4 B ) sin 2 x + ( 4 A + B ) cos 2 x = 17 sin 2 x (A - 4B)\sin 2x + (4A + B)\cos 2x = 17\sin 2x ( A − 4 B ) sin 2 x + ( 4 A + B ) cos 2 x = 17 sin 2 x
{ A − 4 B = 17 4 A + B = 0 ⇒ A = 1 , B = − 4 \left\{ \begin{array}{l}
A - 4B = 17\\
4A + B = 0
\end{array} \right. \Rightarrow A = 1,B = - 4 { A − 4 B = 17 4 A + B = 0 ⇒ A = 1 , B = − 4
So,
y ~ = sin 2 x − 4 cos 2 x \widetilde y = \sin 2x - 4\cos 2x y = sin 2 x − 4 cos 2 x
y = y 0 + y ~ = e − x ( C 1 cos 2 x + C 2 sin 2 x ) + sin 2 x − 4 cos 2 x y = {y_0} + \widetilde y = {e^{ - x}}\left( {{C_1}\cos 2x + {C_2}\sin 2x} \right) + \sin 2x - 4\cos 2x y = y 0 + y = e − x ( C 1 cos 2 x + C 2 sin 2 x ) + sin 2 x − 4 cos 2 x
Answer: y = e − x ( C 1 cos 2 x + C 2 sin 2 x ) + sin 2 x − 4 cos 2 x y = {e^{ - x}}\left( {{C_1}\cos 2x + {C_2}\sin 2x} \right) + \sin 2x - 4\cos 2x y = e − x ( C 1 cos 2 x + C 2 sin 2 x ) + sin 2 x − 4 cos 2 x
#339914 on Dec 2023
Comments
In question1 How to get D =2500−1600=900? can you explain in detail