## Firstorder Firstdegree Linear Differential Equations

16.2. {a) Use the formula for a general solution to solve the following equation. (b) Check your answer.

dt a) Here i/ = 5 and z = 0. Substituting in {16.1), y(t) = e~S5dt(^A + joeSSdtdt^J

Integrating the exponents, / 5 dt = 5t + c, where c can be ignored because it is subsumed under A Thus, y(t) = e~5t(A + /0 dt). And J0dt = k, a constant, which can also be subsumed under A Hence, y(t) = e~5tA = Ae~5t (16Jl b) Taking the derivative of (16.20), dyldt = -5Ae~5t. From (16.19), dyldt = -5y. Substituting y frcm (16.20),

Rearranging to obtain the general format, dy n a Here v = — 3 and z = 0. Substituting in (16.1), y(t) = e-s-3dt^A + J 0es~3dtdt

Substituting f — 3dt = —3t, y(t) = e3t(A + f0dt)= Ae3t. At t = 0, y = 2. Thus, 2 = Aem, A = 2. Substituting, y(f) = 2e3t (16.22)

b i Taking the derivative of (16.22), dy/dt = 6e3t. From (16.21), dy/dt = 3y. Substituting y from (16.22), dy/dt = 3(2e3t) = 6e3t.

Redo Problem 16.2, given a) Here v - 0 and z = 15. Thus, dt y(t) = e~sodt^A +

where / 0 di = A:, a constant. Substituting and recalling that ek is also a constant, y(t) = e~k(A + J 15 ekdtj

where A is an arbitrary constant equal to Ae k or simply c. Whenever the derivative is equal to a constant, simply integrate as in Example 1. b) Taking the derivative of (16.24), dyldt = 15. From (16.23), dy/dt = 15.

IL5. Redo Problem 16.2, given

dt a) Here v = -6, z = 18, and / -6 dt = -6t. Substituting in (16.1)

y(t) = e6t(^A + J 18 e~6tdt where fl8e~6'dt = -3e~6t. Thus, y(t) = e6t(A - 3e~6t) = Ae6t - 3 (16.26)

b) Taking the derivative of (16.26), dyldt = 6Ae6t. From (16.25), dyldt = 18 + 6y. Substituting y from (16.26), dy/dt = 18 + 6(Ae6t - 3) = 6Ae6t.

16.6. Redo Problem 16.2, given

where J -2U**d/ - -5*\ Substituting. y(t) - «•*(/! ~ 50 - A** - 5. At / - 0, y - 10. Thu* 10 = At M* - 5, and /I = 15. Substituting.

6) Hie derivative of (16.28) is dyidl~ -toe 4'. From (16.27), dyfdt« -20 - 4v. Substituting from >«) for y. dy/rf/ - -20 - 4(1*5*-* - 5) - -60e"*.

16.7. Redo Problem 16.2, given dt o) r - 4/. c - 6r. and /4/ifi - 2r. Thus.

Using the substitution method for the remaining integral, let u - 2l*. dw'dt - 4#. and dt - ¿to,'4/.

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