Jim Lambers MAT 285 Spring Semester Practice Exam 2 Solution. y(t) = 5 2 e t 1 2 e 3t.

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1 . Solve the initial value problem which factors into Jim Lambers MAT 85 Spring Semester 06-7 Practice Exam Solution y + 4y + 3y = 0, y(0) =, y (0) =. λ + 4λ + 3 = 0, (λ + )(λ + 3) = 0. Therefore, the roots are λ = and λ = 3. Since the roots are real and distinct, the general solution is y(t) = c e t + c e 3t, which has the derivative y (t) = c e t 3c e 3t. y(0) = c + c =, y (0) = c 3c =. By adding these two equations, we obtain c =, and therefore c = /. Substituting this value into either equation yieilds c = 5/. We conclude that the solution is y(t) = 5 e t e 3t.. Solve the initial value problem y + 4y + 5y = 0, y(0) =, y (0) = 0. λ + 4λ + 5 = 0, which has roots λ, = 4 ± 4 4()(5) = ± = ± i.

2 Since the roots are complex, the general solution is y(t) = c e t cos t + c e t sin t, which has the derivative y (t) = c e t cos t c e t sin t c e t sin t + c e t cos t. y(0) = c =, y (0) = c + c = 0. It follows that c = and c =. We conclude that the solution is y(t) = e t cos t + e t sin t. 3. Use reduction of order to find a second solution of the equation xy y + 4x 3 y = 0, y (x) = sin(x ). Solution First, we divide both sides of the equation by the coefficient of x to obtain y x y + 4x y = 0. Then, this equation has the form y + p(x)y + q(x)y = 0, where p(x) = /x. We assume that the second solution y (x) has the form y (x) = y (x)v(x). Then v(x) is given by v(x) = = = = p(x) dx [y (x)] e dx [sin(x )] e [sin(x )] eln x dx x [sin(x )] dx x dx dx

3 = = sin u du, csc u du u = x = cot u = cot(x ). Neglecting the constant factor, we conclude that a second solution is 4. Solve the initial value problem which has roots y (x) = sin(x ) cot(x ) = sin(x ) cos(x ) sin(x ) = cos(x ). 9y y + 4y = 0, y(0) =, y (0) =. 9λ λ + 4 = 0, λ, = ± ( ) 4(9)(4) (9) Since this is a double root, the general solution is of the form which has the derivative y(t) = c e t/3 + c te t/3, = 3. y (t) = 3 c e t/3 + c e t/3 + 3 c te t/3. y(0) = c =, y (0) = 3 c + c =. It follows that c = and c = ()/3 = 7/3. We conclude that the solution is y(t) = e t/3 7 3 tet/3.

4 5. Find the general solution of the differential equation y + y + y = e t. λ + λ + = 0, which factors into (λ + ) = 0. It follows that the roots are both equal to, and therefore the general solution of the homogeneous equation is y h (t) = c e t + c te t. Now, we need to find a particular solution of the inhomogeneous equation. Since the equation has constant coefficients and the right-hand side g(t) = e t is of an appropriate form, we can use the method of undetermined coefficients. Specifically, g(t) is of the form P n (t)e at. Therefore, the particular solution is of the form y p (t) = t s (A 0 + A t + + A n t n )e t. In this case, n = 0, P 0 (t) =, and a =. Since a is occurs twice as a root of the characteristic equation, s =. Therefore, the particular solution has the form where A 0 is an undetermined coefficient. y p (t) = t A 0 e t We now substitute this form of y p (t) into the ODE. From we obtain which simplifies to It follows that A 0 = and that y p(t) = A 0 (t t )e t, y p(t) = A 0 (t 4t + )e t, A 0 (t 4t + )e t + [A 0 (t t )e t ] + A 0 t e t = e t, We conclude that the general solution is A 0 =. y p (t) = t e t. y(t) = y h (t) + y p (t) = c e t + c te t + t e t.

5 6. Find the general solution of the differential equation t y ty + y = 4t, y (t) = t, y (t) = t. Solution Since this inhomogeneous equation does not have constant coefficients, we cannot use the method of undetermined coefficients. Instead, we can use variation of parameters. First, we divide both sides of the equation by the coefficient of t, to obtain y t y + t y = 4. Then, compute the Wronskian of the homoegeneous solutions y and y, which is W (y, y )(t) = y (t)y (t) y (t)y (t) = t(t ) t (t ) = t t = t. Then, the particular solution is of the form y p (t) = y (t)w (t) + y (t)w (t) where w (t) = w (t) = y (t)g(t) 4t W (y, y )(t) dt = t y (t)g(t) W (y, y )(t) dt = 4t t dt = 4 dt = 4 dt = 4t, dt = 4 ln t. t We conclude that a particular solution is and that the general solution is y p (t) = 4t + 4t ln t, y(t) = y h (t) + y p (t) = c t + c t + 4t ln t. In the general solution, the 4t term has been dropped because it is a solution of the homogeneous equation. 7. A object with a mass of 00 g stretches a spring 5 cm. If the object is set in motion from its equilibrium position with a downward velocity of 0 cm/s, and if there is no damping, determine the position u of the object at any time t. Find the frequency, period, and amplitude of the motion. Solution Since there is no damping and no external force, the motion of the object is modeled by the equation mu + ku = 0,

6 where u is measured in meters, m = 00 g, and the spring constant k satisfies the equation kl mg = 0, where L = 0.05 m and g = 9.8m/s. Therefore, The initial conditions are k = mg L = (00 g)(9.8 m/s ) = 9, 600 g/s m u(0) = 0, u (0) = 0., since the object is set in motion from its equilibrium position with a downward velocity of 0 cm/s, which is 0. m/s. Dividing both sides of the ODE by the coefficient of u yields The characteristic equation is u + 96u = 0. λ + 96 = 0, which has complex roots λ, = ±4i. It follows that the solution is of the form which has a derivative of u(t) = A cos 4t + B sin 4t, u (t) = 4A sin 4t + 4B cos t. u(0) = A = 0, u (0) = 4B = 0., which yields A = 0 and B = /40. Therefore, the solution is The frequency is The period is and the amplitude is ω 0 = u(t) = sin 4t. 40 k 9600 m = 00 = 96 = 4. T = π ω 0 = π 4 = π 7, R = A + B = 40.