Math 135-2, Homework 1

Transcription

1 Mth 35-, Homework Solutions Problem 7. Find the generl solution of ech of the following equtions: ) 6y 8y + y q) y + 4y + 5y r) y + 4y 5y ) 6y 8y +y y e rx y re rx y r e rx 6r e rx 8re rx +e rx 6r 8r +)e rx 6r 8r + 4r ) q) y +4y +5y r 4, 4 y c +c x)e 4 x r e rx +4re rx +5e rx r +4r +5)e rx r +4r +5 r +) + r ±i y c e +i)x +c e i)x c cosx+c 3 sinx)e x

2 r) y +4y 5y r e rx +4re rx 5e rx r +4r 5)e rx r +4r 5 r +5)r ) r, 5 y c e x +c e 5x Problem 7.5 The eqution x y + xy + qy, where nd q re constnts, is clled Euler s equidimensionl eqution. Show tht the chnge of indeendent vrible given by x e z trnsforms it into n eqution with constnt coefficients, nd ly this technique to find the generl solution of ech of the following equtions: ) x y + 3xy + y Some degree of cre is helful here. Let ŷz) yx) ye z ), where I hve voided clling both functions y since they re in fct different functions. ŷ d dz yez )) y e z )e z xy ŷ ŷ d dz y e z )e z ) y e z )e z e z +y e z )e z y x)x +y x)x x y ŷ y x)x ŷ ŷ x y +xy +qy ŷ ŷ +ŷ +qŷ ŷ + )ŷ +qŷ

3 As exected, this eqution hs constnt coefficients. x y +3xy +y ŷ ŷ +3ŷ +ŷ ŷ +ŷ +ŷ ŷ e rz r +r + r ± 4 ± 36 ±3i ŷz) c cos3z)+c sin3z))e z yx) c cos3lnx)+c sin3lnx))e lnx yx) c cos3lnx)+c sin3lnx))x z lnx Note tht this rocess in equivlent to choosing y x r s tril function rther thn ŷ e rz. Problem 8.3 If y x) nd y x) re solutions of nd y + Px)y + Qx)y R x) y + Px)y + Qx)y R x), show tht yx) y x) + y x) is solution of y + Px)y + Qx)y R x) + R x). This is clled the rincile of suerosition. Use this rincile to find the generl solution of b) y + 9y sin3x + 4sinx 6e x + 7x 3 y +Px)y +Qx)y y +y ) +Px)y +y ) +Qx)y +y ) Next, lets tckle the homogeneous roblem y +Px)y +Qx)y )+y +Px)y +Qx)y ) R x)+r x) y +9y y e rt r +9)e rt r ±3i y c cos3x+c sin3x 3

4 Finlly, lets del with the right hnd sides one by one. First sin3x. For this one, note tht we will need to introduce fctor of x. y xsin3x+bxcos3x y sin3x+3xcos3x+bcos3x 3bxsin3x y 6cos3x 9xsin3x 6bsin3x 9bxcos3x y +9y 6cos3x 9xsin3x 6bsin3x 9bxcos3x)+9xsin3x+bxcos3x) 6cos3x 6bsin3x sin3x b 3 y 3 xcos3x Next is 4sinx. y sinx+bcosx y cosx bsinx y sinx bcosx y +9y sinx bcosx)+9sinx+bcosx) 8sinx+8bcosx 4sinx b y sinx Now for 6e x. y e x y e x y 4e x y +9y 3e x 6e x y e x 4

5 Finlly, lets do 7x 3. y x 3 +bx +cx+d y 3x +bx+c y 6x+b y +9y 6x+b)+9x 3 +bx +cx+d) 9x 3 +9bx +9c+6)x+9d+b) 7x 3 3 b c d y 3x 3 x Putting this ll together, y c cos3x+c sin3x 3 xcos3x+ sinx e x +3x 3 x Problem 48. Evlute the integrls in 8), 9), ), ), nd 3). L[] e x dx [ ] e x lim x e x + limit requires > ) L[x] e x xdx [ ) ] e x x ) e x dx e x dx limit requires > ) du e x dx,v x 5

6 L[e x ] e x e x dx e )x dx [ e )x ] lim x e )x + limit requires > ) Let A L[sinx] nd B L[cosx]. A L[sinx] e x sinxdx [ ) ] e x sinx B B L[cosx] ) e x cosxdx e x cosxdx limit requires > ) e x cosxdx [ ) ] e x cosx A ) e x )sinxdx e x sinxdx limit requires > ) du e x dx,v x du e x dx,v x B + )B B + A B + Note tht the lst two could lso be obtined by using e ix cosx + isinx, which roduces simler 6

7 integrl. In tht cse, L[cosx]+iL[sinx] L[cosx+isinx] L[e ix ] i e x e ix dx e i)x dx + + limit requires > ) + i Problem 48. Without integrting, show tht ) L[sinhx], > [ e x e x ] L[sinhx] L L[ex ] L[e x ] + Note tht the Llce trnsforms used require > nd >, leding to >. Problem 48.4 Use the formuls given in the text to find the trnsform of ech of the following functions: d) 4sinxcosx + e x L[4sinxcosx+e x ] L[sinx+e x ] L[sinx]+L[e x ] Problem 48.5 Find function fx) whose trnsform is e) 4 + 7

8 4 + A + B + C + A +)+B +)+C B C i A+B +C A fx) x sinx Problem 49. In ech of the following cses, grh the function nd find its Llce trnsform: ) fx) ux ) where is ositive number nd ux) is the unit ste function defined by { if x < ux) if x b) fx) [x] where [x] denotes the gretest integer x c) fx) x{ [x] sinx if x π d) fx) if x > π ) b) L[ux )] ux )e x dx e x dx+ e x dx [ ] e x 3 4 e 8

9 L[[x])] k [x]e x dx k+ k ke x dx [ k ] k+ e x k k k e k+) + k ) e k k 3 4 e ke k k To evlute this summtion, note tht k k z k z kz k z) k kz k z z) k ke k e e ) L[[x])] e ke k k e e ) e ) ) e e e ) 3 c) d) L[x [x])] L[x] L[[x])] e ) 3 4 L[fx)] fx)e x dx sinxe x dx+ sinxe x dx π e x dx

10 This integrl is similr to one bove. Let A A e x sinxdx [ ) ] π e x sinx B e x cosxdx sinxe x dx nd B ) e x cosxdx sinxe x dx. du e x dx,v x B e x cosxdx [ ) ] π e x cosx e π + e π + A e x sinxdx ) e x sinxdx du e x dx,v x e π + B +)B e π +) B e π +) + A B e π + + Problem 49.4 Show exlicitly tht L[x ] does not exist. L[x ] x e x dx x e x dx+ x e x dx Using x shows tht the second integrl exists. The first integrl is the roblem. Indeed, This integrl diverges by the -test. x e x dx x e dx e x dx

11 Problem 5. Find the Llce trnsforms of b) x )e x L[ x ] 3 L[ x )e x ] + +) 3 Note tht this technique cn lso be used to del with fctors like sinx or cosx. For exmle, Problem 49.d) ws very tedious to do directly. Using the shift rule, this cn be simlified somewht. fx) ux π))sinx sinx Imux π)e ix ) L[fx)] L[sinx Imux π)e ix )] L[fx)] L[sinx] Im L[ux π)e ix ] ) L[fx)] + Im L[ux π)] ) i L[fx)] ) + Im e π i L[fx)] ) + Im i e i)π L[fx)] ) +i + Im + e π cosπ +isinπ) L[fx)] ) +i + Im + e π ) L[fx)] e π L[fx)] e π + +