Math 22B, Homework #8 1. y 5y + 6y = 2e t

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1 Math 22B, Homework #8 3.7 Problem # We find a particular olution of the ODE y 5y + 6y 2e t uing the method of variation of parameter and then verify the olution uing the method of undetermined coefficient. VOP Firt we olve the homogeneou equation uing the characteritic equation r 2 5r + 6 which ha root r 3, 2. a fundamental et of olution for the homogeneou equation i y e 3t and y 2 e 2t we aume that a particular olution ha the form differentiating we get that y p (t) u (t)e 3t + u 2 (t)e 2t y p 3u e 3t + 2u 2 e 2t y p 3(u + 3u )e 3t + 2(u 2 + 2u 2 )e 2t The form of y p come from the contituent equation u e 3t + u 2e 2t () Plugging y p and it derivative back into the ODE we get a econd equation 3u e 3t + 2u 2e 2t 2e t (2) Putting equation () and (2) together we get the ytem [ [ [ e 3t e 2t u 3e 3t 2e 2t u 2 2e t Solving the ytem we obtain [ u u 2 [ 2e 2t 2e t after integrating we get: u e 2t and u 2 2e t Returning to the original form we get y p (t) u (t)e 3t + u 2 (t)e 2t e 2t e 3t + 2e t e 2t e t

2 Math 22B, Homework #8 2 UC Now we can check the anwer above by the following: Aume that y p where A i a contant. Then we get Ae t A and y p e t Ae t 5Ae t + 6Ae t 2e t 2A 2 Problem #5 We ue Variation of Parameter to olve the ODE y + y tan t < t < π 2 y h (t) The characteritic equation i r 2 + which ha root r ±i. the general olution to the homogeneou equation i VOP We aume that y h A co t + B in t y p (t) u (t) co t + u 2 (t) in t Then the derivative of y p have the form y p u in t + u 2 co t y p (u 2 u ) co t (u + u 2 ) in t The form of y p come from the contituent equation u co t + u 2 in t () Plugging y p and it derivative back into the ODE we get another equation u in t + u 2 co t tan t (2) Putting equation () and (2) together we get the ytem [ [ [ co t in t u in t co t u 2 tan t Which after olving yield [ u u 2 [ ec t + co t in t u (t) Now we have u (t) ec t + co t thu u (t) ( ec t + co t)dt ec tdt + co tdt The econd integral i eay, but to do the firt we write ec t co t co t co t co t co t in 2 t

3 Math 22B, Homework #8 3 So ( ec tdt Now, let u in t and du co tdt, then ( ) co t du in 2 dt t u [ 2 2 co t in 2 t ) dt du + u + du u 2 ln + u u So finally we get ( ) co t ec tdt in 2 dt t 2 ln + in t in t u (t) in t 2 ln + in t in t u 2 (t) Thankfully, the expreion for u 2 i much eaier. We have u 2 in t thu u 2 (t) co t y p (t) Now we aumed that y p (t) u (t) co t + u 2 (t) in t So [ y p (t) in t 2 ln + in t in t co t + [ co t in t y p (t) 2 ln + in t in t co t y(t) Now for the general olution y(t) y h (t) + y p (t), thu y(t) A co t + B in t 2 ln + in t in t co t Problem #6 Given that y e t and y 2 t form a fundamental et of olution for ( t)y + ty y we find a particular olution for Aume that Then the derivative of y p become with the contituent equation ( t)y + ty y 2(t ) 2 e t y p (t) u e t + u 2 t y p u e t + u 2 y p (u + u )e t + u 2 u e t + u 2t ()

4 Math 22B, Homework #8 4 Plugging y p and it derivative into the ODE we get Putting () and (2) together we have [ e t t [ u e t u 2 u e t + u 2 2( t)e t (2) [ 2( t)e t Inverting give [ u u 2, after integrating, we get [ 2te 2t 2e t u (t) te 2t + 2 e 2t and u 2 (t) 2e t y p (t) u e t + u 2 t (te 2t + 2 e 2t ) e t + ( 2e t )t 2 e t te t Problem #2 Let y(t) be any olution to the IVP L[y y + p(t)y + q(t)y g(t) ; y(t ) y ; y (t ) y By Theorem 3.2. there exit a unique function u(t) which i a olution to the IVP L[u u + p(t)u + q(t)u ; u(t ) y ; u (t ) y Then let v(t) y(t) u(t). We have L[v L[y u L[y L[u g(t) g(t) and v(t ) y(t ) u(t ) y y ; v (t ) y (t ) u (t ) y y y(t) v(t) + u(t) i a olution to the original IVP. 6. Problem #5 (a) Find L[t L[t te t dt [ te t 2 e t 2 (b) Find L[t 2 L[t 2 t 2 e t dt [ t2 e t 2 2 e t 2 3 e t 2 3

5 Math 22B, Homework #8 5 (c) Find L[t n. L[t n t n e t dt n! n+ The reaon i that all the term in the integral are of the form t k e t for the lat term. The coefficient of the lat term i n!. n+ k except Problem #9 Find L[e at coh bt ( ) e L[e at coh bt L [e at bt + e bt 2 2 (L[e(a+b)t + L[e (a b)t ) 2 (L[e(a+b)t + L[e (a b)t ) [ 2 (a + b) + a (a b) ( a) 2 b 2 L[e at a coh bt ( a) 2 b 2 Problem #3 Find L[e at in bt L[e at in bt [ L[e (a+ib)t L[e (a ib)t [ 2i 2i (a + ib) (a ib) Problem #8 Find L[t n e at 6.2 Let S a. Then Problem # Solve the IVP L[t n e at uing a Laplace tranform. t n e (a )t dt L[e at in bt t n e at e t dt L[t n e at y y 6y ; b ( a) 2 + b 2 t n e (a )t dt t n e St dt n! S n+ n! ( a) n+ n! ( a) n+ y() ; y () L[y y 6y ( 2 Y () y() y ()) (Y () y()) 6Y () Y () y(t) 5 e3t e 2t b ( a) 2 + b 2

6 Math 22B, Homework #8 6 Problem #4 Solve the IVP y 4y + 4y ; y() ; y () uing a Laplace tranform. L[y 4y + 4y ( 2 Y () y() y ()) 4(Y () y()) + 4Y () Y () 3 ( 2) 2 2 ( 2) 2 ( 2) 2 Y () 2 ( 2) 2 y(t) e 2t te 2t ( t)e 2t Problem #6 Solve the IVP y + 2y + 5y ; y() 2 ; y () uing a Laplace tranform. L[y + 2y + 5y ( 2 Y y() y ()) + 2(Y y()) + 5Y Y () Problem #8 Solve the IVP (2 + 2) + ( ) ( + ) ( + ) y(t) 2e t co 2t + 2 e t in 2t y iv y ; y() ; y () ; y () ; y () uing a Laplace tranform. L[y iv y ( 4 Y 3 y() 2 y () y () y ()) Y Y () y(t) coh t 2 Problem #2 Solve the IVP uing a Laplace tranform. y 2y + 2y co t ; y() ; y () L[y 2y + 2y ( 2 Y y() y ()) 2(Y y()) + 2Y 2 + L[co t

7 Math 22B, Homework #8 7 ( ) ( ) 2 Y () ( } {{ } 2 + ) ( ) } {{ } () (2) We can deal with the two part () and (2) eperately ince L i a linear operator. Furthermore () correpond to the homogeneou equation and (2) to a particular olution to the given ODE. () We have Y h () ( ) 2 + ( ) 2 + ( ) 2 + y h (t) e t co t e t in t (2) Here we have Now we aume that Y p () ( 2 + ) ( ) ( 2 + ) ( ) A + B C + D 2 + Solving thi we get A 5 ; B 4 5 ; C 5 ; D 2 5. Y p () [ [ ( ) ( ) And finally Y p (t) 5 ( ) ( ) ( ) Hence y p (t) 5 et co t et in t + 5 co t 2 5 in t Finally if we put () and (2) together we get y(t) y h (t) + y p (t) 4 5 et co t 2 5 et in t + 5 co t 2 5 in t Proble #23 Solve the IVP y + 2y + y 4e t ; y() 2 ; y () uing a Laplace tranform. L[y + 2y + y ( 2 Y y() y ()) + 2(Y y()) + Y 4 + L[4e 4t

8 Math 22B, Homework #8 8 And finally Y () ( + )( ) Y () + + ( + ) ( + ) 3 y(t) 2e t + te t + 2t 2 e t 2( + ) ( + ) 2 ( + ) 3

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