springps06_soln

springps06_soln - 18.03 PS6 Solutions—Spring 09 1a x = e...

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Unformatted text preview: 18.03 PS6 Solutions—Spring 09 1a. x = e t = ⇒ dy dt = dx dt dy dx = e t dy dx . Hence, dy dx = e − t dy dt and dy dx − parenleftbigg 3 x parenrightbigg y = x 2 = ⇒ e − t dy dt − 3 e − t y = e 2 t = ⇒ dy dt − 3 y = e 3 t . ( ∗ ) The characteristic polynomial p ( r ) = r − 3 is zero when r = 3, so using ERF ′ , y = te 3 t /p ′ (3) = te 3 t is a particular solution. The homogeneous ODE dy dt − 3 y = 0 has solu- tion y = ce 3 t , meaning that the general solution to ( ∗ ) is y = te 3 t + ce 3 t = x 3 ln x + cx 3 . 1b. From part (a), df dx = e − t df dt for any function f . Let f = dy dx = e − t dy dt . Then d 2 y dx 2 = df dx = e − t df dt = e − t d dt parenleftbigg e − t dy dt parenrightbigg = e − t parenleftbigg e − t d 2 y dt 2 − e − t dy dt parenrightbigg = e − 2 t parenleftbigg d 2 y dt 2 − dy dt parenrightbigg Alternatively, this same calculation may be written using operator notation: d dx = e − t d dt implies d 2 y dx 2 = d dx parenleftbigg dy dx parenrightbigg = e − t d dt parenleftbigg e − t dy dt parenrightbigg = e − t parenleftbigg e − t d 2 y dt 2 − e − t dy dt parenrightbigg = e − 2 t parenleftbigg d 2 y dt 2 − dy dt parenrightbigg Hence the ODE is x 2 d 2 y dx 2 + ax dy dx + by = e 2 t e − 2 t parenleftbigg d 2 y dt 2 − dy dt parenrightbigg + ae t e − t dy dt + by = d 2 y dt 2 + ( a − 1) dy dt + by = 0 which has characteristic polynomial p ( r ) = r 2 + ( a − 1) r + b . Let Δ = ( a − 1) 2 − 4 b be the discriminant. When Δ > 0, there are two distinct real roots r 1 = (1 − a + √ Δ) / 2 and r 2 = (1 − a − √ Δ) / 2, so the solution is y = c 1 e r 1 t + c 2 e r 2 t = c 1 x r 1 + c 2 x r 2 . When Δ < 0, there are two distinct complex roots u + iw and u − iw , where u = (1 − a ) / 2 and w = radicalbig | Δ | / 2. The solution is y = c 1 e ut cos( wt ) + c 2 e ut sin( wt ) = c 1 x u cos( w ln x ) + c 2 x u sin( w ln x ) . When Δ = 0, there is a double root r = (1 − a ) / 2, so the solution is y = c 1 e rt + c 2 te rt = c 1 x r + c 2 x r ln x . 1c. y = x r = ⇒ x 2 y ′′ + axy ′ + by = x 2 r ( r − 1) x r − 2 + axrx r − 1 + bx r = [ r ( r − 1)+ ar + b ] x r = [ r 2 + ( a − 1) r + b ] x r = 0 = ⇒ r 2 + ( a − 1) r + b = 0 . When Δ > 0, so that r 1 and r 2 are distinct real roots as in (b), y = c 1 x r 1 + c 2 x r 2 is the general solution. When Δ < 0, so that u ± iw are distinct complex roots, we could say that y = c 1 x u + iw + c 2 x u − iw is the general solution, but we clarify further the meaning of x r for complex r . For x > 0, we define x r = e r ln x . Then x u + iw = e ( u + iw ) ln x = e u ln x e iw ln x = x u e iw ln x and similarly x u − iw = x u e − iw ln x . Taking the real and imaginary parts of these complex solutions, we get the general real solution y = c 1 x u cos( w ln x ) + c 2 x u sin( w ln x ) ....
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This note was uploaded on 05/06/2010 for the course 18 18.03 taught by Professor Unknown during the Spring '09 term at MIT.

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springps06_soln - 18.03 PS6 Solutions—Spring 09 1a x = e...

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