415am2 - Solution to Test 2 Math 415A Tanveer 1a Find...

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Unformatted text preview: Solution to Test 2, Math 415A, Tanveer 1a. Find solution y ( t ) satisfying the initial value problem: y ′′ + 4 y ′ + 4 y = 0 , y (0) = 0 , y ′ (0) = 1 Solution: Since constant coefficient, homogeneous solution, we look at the characterestic equation r 2 +4 r +4 = 0, or ( r +2) 2 = 0 or r =- 2 twice. Since roots are coincident, two independent solutions are e − 2 t and te − 2 t . So, y ( t ) = c 1 e − 2 t + c 2 te − 2 t Since y (0) = 0, implies c 1 = 0. Since y ′ ( t ) = c 2 e − 2 t- 2 c 2 te − 2 t , y ′ (0) = 1 = c 2 . Therefore, solution to IVP y ( t ) = te − 2 t 1b. Find form of the particular solution in each of the two cases below ( Note: not asking to find the coefficients): i. y ′′ + 4 y ′ + 4 y = e − 2 t (1 + t 2 ) Solution: Since r =- 2 is root of the characteristic equation twice, it follows that we must have y p = t 2 e − 2 t ( A t 2 + A 1 t + A 2 ) ii. y ′′ + y = e t sin( t ) + t Solution: Since 1 + i is not a root of characteristic equation r 2 + 1 = 0 (which has roots ± i ), it follows y p, 1 = A e t sin t + B e t cos t is a particular solution to y ′′ + y = e t sin( t ) Again since r = 0 is not a root of the characteristic equation r 2 +1 = 0, it follows that particular solution to y ′′ + y = t is of the form y p, 2 = A 1 t + B 1 . So particular solution to the problem....
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415am2 - Solution to Test 2 Math 415A Tanveer 1a Find...

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