Exam 17: Second-Order Differential Equations

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Solve the differential equation using the method of variation of parameters. Select the correct answer. yttyt=e2xy ^ { tt } - y ^ { t } = e ^ { 2 x }

(Multiple Choice)
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Graph the particular solution and several other solutions. 2ytt+3yt+y=2+cos2x2 y ^ { tt } + 3 y ^ { t } + y = 2 + \cos 2 x

(Essay)
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Solve the differential equation. Select the correct answer. 25ytt+y=025 y ^ { tt } + y = 0

(Multiple Choice)
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Solve the boundary-value problem, if possible. Select the correct answer. ytt+16yt+64y=0,y(0)=0,y(1)=5y ^ { tt } + 16 y ^ { t } + 64 y = 0 , y ( 0 ) = 0 , y ( 1 ) = 5

(Multiple Choice)
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A spring with a mass of 2 kg2 \mathrm {~kg} has damping constant 14 , and a force of 3.6 N3.6 \mathrm {~N} is required to keep the spring stretched 0.3 m0.3 \mathrm {~m} beyond its natural length. The spring is stretched 1 m1 \mathrm {~m} beyond its natural length and then released with zero velocity. Find the position x(t)x ( t ) of the mass at any time tt .

(Short Answer)
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Use power series to solve the differential equation.. (x2+1)ytt+xyty=0\left( x ^ { 2 } + 1 \right) y ^ { t t } + x y ^ { t } - y = 0

(Short Answer)
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A spring with a 3kg3 - \mathrm { kg } mass is held stretched 0.9 m0.9 \mathrm {~m} beyond its natural length by a force of 30 N30 \mathrm {~N} . If the spring begins at its equilibrium position but a push gives it an initial velocity of 1 m/s1 \mathrm {~m} / \mathrm { s } , find the position x(t)x ( t ) of the mass after tt seconds.

(Short Answer)
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Solve the differential equation. Select the correct answer. ytt8yt+25y=0y ^ { tt } - 8 y ^ { t } + 25 y = 0

(Multiple Choice)
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Find ff by solving the initial value problem. ftt(x)=8x2+12x+4;f(1)=4,ft(1)=6f ^ {tt } ( x ) = 8 x ^ { 2 } + 12 x + 4 ; \quad f ( - 1 ) = - 4 , \quad f ^ { t } ( - 1 ) = - 6

(Short Answer)
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Suppose a spring has mass MM and spring constant kk and let ω=k/M\omega = \sqrt { k / M } . Suppose that the damping constant is so small that the damping force is negligible. If an external force F(t)=4F0cos(ωt)F ( t ) = 4 F _ { 0 } \cos ( \omega t ) is applied (the applied frequency equals the natural frequency), use the method of undetermined coefficients to find the equation that describes the motion of the mass.

(Multiple Choice)
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The solution of the initial-value problem x2ytt+xyt+x2y=0,y(0)=1,yt(0)=0x ^ { 2 } y ^ { tt } + x y ^ {t } + x ^ { 2 } y = 0 , y ( 0 ) = 1 , y ^ { t } ( 0 ) = 0 is called a Bessel function of order 0 . Solve the initial - value problem to find a power series expansion for the Bessel function.

(Multiple Choice)
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Solve the boundary-value problem, if possible. ytt+5yt50y=0,y(0)=0,y(2)=1y ^ { tt } + 5 y ^ { t } - 50 y = 0 , y ( 0 ) = 0 , y ( 2 ) = 1

(Short Answer)
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Solve the differential equation using the method of undetermined coefficients. ytt4yt+5y=8exy ^ { tt } - 4 y ^ { t } + 5 y = 8 e ^ { - x }

(Short Answer)
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Solve the initial-value problem. ytt+81y=0,y(π9)=0,yt(π9)=6y ^ { tt } + 81 y = 0 , y \left( \frac { \pi } { 9 } \right) = 0 , y ^ { t } \left( \frac { \pi } { 9 } \right) = 6

(Short Answer)
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Use power series to solve the differential equation. ytt=36yy ^ {t t } = 36 y

(Multiple Choice)
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Solve the differential equation. 8ytt+yt=08 y ^ { t t } + y ^ { t } = 0

(Short Answer)
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A series circuit consists of a resistor R=96ΩR = 96 \Omega , an inductor with L=8HL = 8 \mathrm { H } , a capacitor with C=0.00125 FC = 0.00125 \mathrm {~F} , and a generator producing a voltage of E(t)=48cos(10t)E ( t ) = 48 \cos ( 10 t ) . If the initial charge is Q=0.001CQ = 0.001 \mathrm { C } and the initial current is 0 , find the charge Q(t)Q ( t ) at time tt .

(Short Answer)
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A spring with a 16kg16 - \mathrm { kg } mass has natural length 0.8 m0.8 \mathrm {~m} and is maintained stretched to a length of 1.21.2 m\mathrm { m } by a force of 19.6 N19.6 \mathrm {~N} . If the spring is compressed to a length of 0.4 m0.4 \mathrm {~m} and then released with zero velocity, find the position x(t)x ( t ) of the mass at any time tt .

(Multiple Choice)
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Solve the initial-value problem. ytt2yt24y=0,y(1)=4,yt(1)=6y ^ {t t } - 2 y ^ { t } - 24 y = 0 , y ( 1 ) = 4 , y ^ { t } ( 1 ) = 6 .

(Short Answer)
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Solve the differential equatic 25ytt+y=025 y ^ { tt } + y = 0

(Short Answer)
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