hw2 - Homework II 2.1.14. y + 2 y = te- 2 t , y (1) = 0...

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Unformatted text preview: Homework II 2.1.14. y + 2 y = te- 2 t , y (1) = 0 This is a first order linear ODE, μ ( t ) = exp R 2 dt = e 2 t . Note that the last equality is a slight lie, since it ignores the constant of integration, but any constant will work. e 2 t ( y + 2 y ) = te- 2 t e 2 t = t d dt ( ye 2 t ) = t ye 2 t = t 2 2 + c Plugging in y (1) = 0, we have 0 = 1 2 + c , so c =- 1 2 . Thus, y ( t ) = e- 2 t t 2- 1 2 . 2.1.16. y + 2 t y = cos t t 2 , y ( π ) = 0, t > This is a first order linear ODE, μ ( t ) = exp Z 2 t dt = e 2 ln | t | = | t 2 | = t 2 t 2 y + 2 t y = cos t d dt ( t 2 y ) = cos t t 2 y = sin t + c As y ( π ) = 0, we know 0 = 0 + c , so c = 0. Thus y ( t ) = 1 t 2 sin t . 2.1.18. ty + 2 y = sin( t ), y π 2 = 1. This is a first order linear ODE, but we must divide by t before we can calculate μ ( t ). So we have the ODE y + 2 t y = sin( t ) t . Note that p ( t ) is the same as in 2.1.16, so μ ( t ) = t 2 as before. Thus t 2 y + 2 t y = t 2 sin( t ) t = t sin( t ). So d dt ( t 2 y ) = t sin t . We can integrate the right hand side using integration by parts: t 2 y = Z t sin tdt =- t cos t + Z cos tdt =- t cos t + sin t + c Plugging in the initial condition, we find π 2 2 · 1 =- π 2 · 0 + 1 + c , so c = π 2 4- 1. Thus y ( t ) =- cos t t + sin t t 2 + π 2- 4 4 t 2 2.1.20. Consider ty + ( t + 1) y = t , y (ln 2) = 1, t > 0. Dividing by t , we have y + t + 1 t y = 1. This is a first order linear ODE, μ ( t ) = exp Z t + 1 t dt = exp Z 1 + 1 t dt = e t +ln | t | = | t | e t = te t (The last equality is because t > 0.) te t y + t + 1 t y = te t , So d dt ( te t y ) = te t ....
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This note was uploaded on 07/17/2008 for the course MATH 415 taught by Professor Costin during the Fall '07 term at Ohio State.

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hw2 - Homework II 2.1.14. y + 2 y = te- 2 t , y (1) = 0...

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