Explanation Result used: Suppose the auxiliary equation is in the form m y ″ + c y ′ + k y = 0 , then Case ( i ) : If the auxiliary equation has distinct real roots λ 1 , and λ 2 of y = e λ x , then c 2 > 4 m k , that is, overdamping. Case ( ii ) : If the auxiliary equation has real double root λ = a of y = e λ x , then c 2 = 4 m k , that is, critical damping. Case ( iii ) : If the auxiliary equation has complex conjugate root λ 1 = a + i ω , λ 2 = a − i ω of y = e λ x , then then c 2 < 4 m k , that is, underdamping”. Calculation: It is known that, the damped mass-spring system can be modeled by m y ″ + c y ′ + k y = 0 (1) Compute the auxiliary equation of the above ODE as follows. Let y = e λ x . That implies, the first and second derivative of y is y ′ = λ e λ x and y ″ = λ 2 e λ x , respectively. Substitute y ″ = λ 2 e λ x and y = e λ x in (1) and simplify further. m λ 2 e λ x + c λ e λ x + k e λ x = 0 e λ x ( m λ 2 + c λ + k ) = 0 m λ 2 + c λ + k = 0 λ 2 + c m λ + k m = 0 That is, the characteristic equation is λ 2 + c m λ + k m = 0 . Compute the roots of the above characteristic equation as follows. Substitute a = 1 and b = c m and c = k m in λ = − b ± b 2 − 4 a c 2 a . λ = − c m ± ( c m ) 2 − 4 ( 1 ) ( k m ) 2 ( 1 ) = − c m ± c 2 m 2 − 4 k m 2 = − c 2 m ± 1 2 c 2 − 4 m k m 2 = − c 2 m + 1 2 m c 2 − 4 m k and − c 2 m − 1 2 m c 2 − 4 m k = − α + β and − α − β [ Let α = c 2 m and β = 1 2 m c 2 − 4 m k ] From the above calculation, it is observed that the roots of λ 2 + c m λ + k m = 0 is real and distinct. Note that, here the value of c 2 > 4 m k , that is, overdamping. Thus, the general solution will be in the form y = c 1 e − ( α − β ) t + c 2 e − ( α + β ) t . In order to check whether the given initial conditions satisfy the differential equation, Solve the system of equations. y 0 = y ( 0 ) = c 1 e − ( α − β ) ⋅ 0 + c 2 e − ( α + β ) ⋅ 0 = c 1 + c 2 Let y 0 = c 1 + c 2 be equation (1). It is known that, the velocity is the rate of change of displacement with respect to time. That is, v = d y d t . Compute the value of v 0 . v 0 = d d t ( y 0 ) = − ( α − β ) c 1 e − ( α − β ) ⋅ 0 − ( α + β ) c 2 e − ( α + β ) ⋅ 0 = − ( α − β ) c 1 − ( α + β ) c 2 For x = α + β , y 0 ( x ) = c 1 x + c 2 x becomes, y 0 ( α + β ) = c 1 ( α + β ) + c 2 ( α + β )

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ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 2.4, Problem 11P

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To show: The ODE y(t)=c1e−(α−β)t+c2e−(α+β)t will satisfy initial condition y(0)=y0 and v(0)=v0 where the value of c1=[(1+αβ)y0+v0β]2 and c2=[(1−αβ)y0−v0β]2.

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Chapter 2.4, Problem 10P

Chapter 2.4, Problem 12P

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The ODE y ( t ) = c 1 e − ( α − β ) t + c 2 e − ( α + β ) t will satisfy initial condition y ( 0 ) = y 0 and v ( 0 ) = v 0 where the value of c 1 = [ ( 1 + α β ) y 0 + v 0 β ] 2 and c 2 = [ ( 1 − α β ) y 0 − v 0 β ] 2 .

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To show: The ODE y(t)=c1e−(α−β)t+c2e−(α+β)t will satisfy initial condition y(0)=y0 and v(0)=v0 where the value of c1=[(1+αβ)y0+v0β]2 and c2=[(1−αβ)y0−v0β]2.

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