Equations and linear algebra thomas honold first

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Equations and Linear Algebra Thomas Honold First-Order Equations First-Order Linear Equations Solutions cont’d 9 y = x / y This ODE can be written as y y + x = 0 and integrated to yield y 2 2 + x 2 2 = C = C 2 R . The solutions are therefore the half-circle parametrizations y ( x )= ± radicalbig r 2 x 2 , x ( r , r ) ( with r > 0 ) . Every IVP y = x / y y ( x 0 )= y 0 negationslash = 0 has a unique solution, as is easily seen. Alternatively, we can rewrite y = d y dx = x y as x d x + y d y = 0 , which is exact with anti-derivative f ( x , y )= x 2 + y 2 2 . This gives whole-circles centered as ( 0 , 0 ) , which are the contours of f , as implicit solutions. The exceptional role of the x -axis, visible in the original explicit ODE, has gone away.
Math 286 Differential Equations and Linear Algebra Thomas Honold First-Order Equations First-Order Linear Equations Solutions cont’d 10 y ′′ + y = 0 Two particular solutions of y ′′ = y are y 1 ( t )= cos t and y 2 ( t )= sin t (with domain R ). For A , B R , since ( A cos t + B sin t ) ′′ = A ( cos t ) ′′ + B ( sin t ) ′′ = A cos t B sin t = ( A cos t + B sin t ) , we obtain further solutions y ( t )= A cos t + B sin t (also with domain R ). Since y ( 0 )= A cos 0 + B sin 0 = A , y ( 0 )= A sin 0 + B cos 0 = B , there is exactly one solution of any IVP y ′′ = y y ( 0 )= A y ( 0 )= B ( A , B R ). Similarly, one can show that the IVP y ′′ = y , y ( t 0 )= y 0 , y ( t 0 )= y 1 ( t 0 , y 0 , y 1 R ) . has exactly one solution of the said form y ( t )= A cos t + B sin t (Exercise). This means that the graphs of the maps t mapsto→ ( y ( t ) , y ( t ) ) , or traces of the curves t mapsto→ ( t , y ( t ) , y ( t ) ) , with y ( t )= A cos t + B sin t , A , B R , partition the space R 3 . (Can you imagine that?).
Math 286 Differential Equations and Linear Algebra Solutions cont’d 10 (con’t)
Math 286 Differential Equations and Linear Algebra Thomas Honold

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