pset_2sol

# pset_2sol - ACM 95/100b Problem Set 2 Solutions Faisal...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ACM 95/100b Problem Set 2 Solutions Faisal Amlani January 24, 2009 Problem 1 Solve the following second order ODE using variation of parameters y ′′ − 5 y ′ + 6 y = cos( x ) Using your general solution, determine the solution to the initial value problem y (0) = 1 y ′ (0) = 0 SOLUTION : We first solve the homogeneous equation y ′′ − 5 y + 6 y = 0 . Assume the form y = exp( αx ) for the solution to the homogeneous equation (i.e. characteristic equation) = ⇒ α 2 − 5 α + 6 = 0 . Clearly, solutions to this are α 1 = 2 , α 2 = 3 . Thus, two solutions to the homogeneous ODE are (ignore the constants without loss of generality): y 1 ( x ) = exp(2 x ) , y 2 ( x ) = exp(3 x ) . So want to find a particular solution y p such that the solution to the inhomogeneous ODE is given as y ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) + y p ( x ) for y 1 = exp(2 x ) , y 2 = exp(3 x ) solutions to the homogeneous ODE. By variation of parameters, we want y p as y p ( x ) = u 1 ( x ) y 1 ( x ) + u 2 ( x ) y 2 ( x ) . 1 We then have that (from class) that the solution for the inhomogeneous right hand side r ( x ) = cos( x ) y p ( x ) = − y 1 ( x ) integraldisplay x y 2 ( t ) r ( t ) / W(t) dt + y 2 ( x ) integraldisplay x y 1 ( t ) r ( t ) / W(t) dt where the Wronskian W is given by W(t) = det vextendsingle vextendsingle vextendsingle vextendsingle y 1 ( t ) y 2 ( t ) y ′ 1 ( t ) y ′ 2 ( t ) vextendsingle vextendsingle vextendsingle vextendsingle = det vextendsingle vextendsingle vextendsingle vextendsingle exp(2 x ) exp(3 x ) 2 exp(2 x ) 3 exp(3 x ) vextendsingle vextendsingle vextendsingle vextendsingle = exp(5 x ) . Thus, plugging this and y 1 = exp(2 x ) , y 2 = exp(3 x ) , r ( t ) = cos( t ) we can integrate by parts to find that y p ( x ) = − exp(2 x ) integraldisplay x exp( − 2 t ) cos( t ) dt + exp(3 x ) integraldisplay x exp( − 3 t ) cos( t ) dt = − exp(2 x ) bracketleftbigg − 2 5 exp( − 2 x ) cos( x ) + 1 5 exp( − 2 x ) sin( x ) bracketrightbigg + + exp(3 x ) bracketleftbigg − 3 10 exp( − 3 x ) cos( x ) + 1 10 exp( − 3 x ) sin( x ) bracketrightbigg...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

pset_2sol - ACM 95/100b Problem Set 2 Solutions Faisal...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online