Mah Rahman Exam Review Soluions () Consider he IVP: ( 4)y 3y + 4y = ; y(3) = 0, y (3) =. (a) Please deermine he longes inerval for which he IVP is guaraneed o have a unique soluion. Soluion: The disconinuiies are a = 0, 4. The iniial condiion is = 3, so (0, 4). (b) Please compue he Wronskian using Abel s heorem. Soluion: ( W (y, y ) = C exp 3 ) d = C exp (3 ln 4 ). 4 Noice ha 4 < 0 for (0, 4), so 4 = 4, hen W = C exp ( ln(4 ) 3) = C(4 ) 3. () Consider he IVP: 4y + y + 9y = 0; y(0) =, y (0) = α. Soluion: 4r + r + 9 = 0 r + (/4)r + (3 / ) = 0 (r + 3/) = 0 r = 3/ y = (c + c )e 3/. The firs iniial condiions give us y(0) = c = y = ( + c )e 3/. For he second iniial condiion les ake he derivaive, y = c e 3/ 3 ( + c )e 3/, so y (0) = c + 3/ = α c = α 3/, hen he soluion is y = [ + (α 3/)]e 3/. (a) For wha α does he soluion change signs a = /? Soluion: A change of sign always happens around he ime he curve his zero, = ( y = 7 4 + α ) e 3/4 = 0 α = 7 4 α = 7. (b) How many imes does his soluion (for he α above) change signs for > 0? Soluion: Same idea as above, α = 7 y = ( + )e 3/ = 0 =. Since we only have one soluion for, y will change signs only once. (3) One soluion o y 3y + 4y = 0 is y =. Please find he oher soluion. Soluion: Le y = v()y y = v y + vy y = v y + v y + vy. Plugging his ino he ODE gives vy + v y + vy 3v y 3vy + 4vy = v y + ( y 3y )v + 0 ( y 3y + 4y )v = 0. Now le u = v, hen y u = (3y y )u 4 u = (3 3 4 3 )u u = u du d u = ln u = ln + c u = k y = v()y = k ln + c y = ln. v = k ln + c
(4) One soluion o y + 3y y = 0 is y = /. Please find he oher soluion. Soluion: Le y = v()y y = v y + vy y = v y + v y + vy. Plugging his ino he ODE gives v y + 4 v y + vy + 3v y + 3vy vy = v y + (4 y + 3y )v + 0 ( y + 3y y )v = 0. Le u = v, u y = (4 y + 3y )u u = u du u = d ln u = ln + c u = k v = k 3/ + c y = k / + c y =. (5) Please use he mehod of undeermined coefficiens o find he form of he paricular soluion (WITHOUT SOLVING FOR CONSTANTS) of he following ODEs. (a) y + 5y + 6y = + e 3 + e + e 3 cos Soluion: The characerisic soluion is y c = c e + c e 3. We ake a guess a he paricular soluion using our forcing funcion y p? = a 0 + a + ke 3 + (b 0 + b )e + e 3 [A cos + A sin ]. Noice ha here is a repea wih e 3 and e, so our paricular soluion becomes y p = a 0 + a + ke 3 + (b 0 + b )e + e 3 [A cos + A sin ]. (b) y + 3y + y = e ( + ) sin() + 3e cos + 4e. Soluion: The characerisic soluion is y c = c e + c e. We ake a guess a our paricular soluion, bu as we ll see, we won have any repeas, y p = (a 0 + a + a )e [A cos + A sin ] + e [B cos + B sin ] + b 0 e. (6) Please find he general soluion of he ODE: y + 4y + 4y = e ; > 0 Soluion: The characerisic soluion is y c = (c + c )e, so our wo soluions are y = e and y = e. The Wronskian is W (y, y ) = e e e e e = e 4. Then we use our formula for variaion of parameers y f() y = y W (y, y ) d + y y f() e e e d = e d + e e d W (y, y ) e 4 e 4 = e [ln + c 3 ] + e [ + c 4 ]. Since hey ask for he general soluion, here s no need o simplify i any furher.
(7) Consider he ODE y + y + y = cos. (a) Please find he general soluion. Soluion: The characerisic polynomial is r = ( ± 4 8) = ± i, hen y c = e [A cos + A sin ]. We ll see ha here won be any repeas, so he paricular soluion is y p = B cos + B sin. y p = B sin + B cos y p = B cos B sin. Plugging i ino he ODE gives B cos B sin B sin + B cos + B cos + B sin = cos (B + B ) cos + (B B ) sin = cos B = B B = 5, B = 5 y = e [A cos + A sin ] + 5 cos + 5 sin. (b) Wha happens o he soluion as? Soluion: As, y cos + sin 5 5 (8) Consider he ODE y y + y =. (a) One soluion o he homogeneous ODE is y =. Use reducion of order o find he oher soluion y. Soluion: Le y = v()y y = v y + vy y = v y + v y + vy. Plugging his ino he ODE gives v y + 4 v y + vy v y vy + vy = v y + (4 y y )v + ( 0 y y + y )v = 0. Le u = v y u = (y 4 y )u 3 u = 3 u du d u = 3 ln u = 3 ln + c u = k 3/ v = k / + c y = k / + c y = /. (b) Use he characerisic soluion y c = c y + c y o find he general soluion o he full ODE. Soluion: The forcing funcion is f() = / (i.e. sandard form) and Wronskian is W (y, y ) = / / = / Then we plug ino our formula o ge y = y y f() W (y, y ) d + y = [ / + c 3 ] + / [ + c 4 ] y f() / (/) / d = W (y, y ) d + / / (/) / d / Again, since only he general soluion was required here is no need o simplify any furher.
(9) Please solve he IVP: y + 4y = 6 sin(4); y(0) = y (0) = 0. Soluion: The characerisic soluion is y c = A cos + A sin, and we ll see ha here are no repeas so he paricular soluion is y p = B cos 4 + B sin 4 y p = 4B sin 4 + 4B cos 4 y p = 6B cos 4 6B sin 4 Plugging i ino he ODE gives 6B cos 4 6B sin 4 + 4B cos 4 + 4B sin 4 = B cos 4 B sin 4 = 6 sin 4 B = 0, B = y = A cos + A sin sin 4. The firs iniial condiion gives y(0) = A = 0, and he second iniial condiion gives y (0) = A = 0 A =, hen our soluion is y = sin sin 4 (0) Consider he IVP y 3y 4y = + ; y(0) = 3, y (0) = 0. (a) Please find he soluion o he IVP. Soluion: The characerisic soluion is y c = c e 4 +c e, and we ll see ha here are no repeas so our paricular soluion is y p = a + a 0 y p = a y p = 0. Plugging ino he ODE gives 3a 4a 4a 0 = + a = 4 a 0 = 5 6 y = c e 4 + c e 4 5 6. The firs iniial condiion gives us y(0) = c + c 5/6 = 3 c + c = 53/6. The derivaive of he soluion is y = 4c e 4 c e /4, hen he second iniial condiion gives us y (0) = 4c c /4 = 0 c c /4 = /6. Then we ge c = 3/5 and c = 57/80 and our soluion becomes y = 57 80 e4 + 3 5 e 4 5 6. (b) Wha happens o he soluion as? Soluion: As, y. () A mass weighing / lb (i.e. mass = /64lb s /f) sreches a spring / f. (a) Suppose he sysem has no damping. The mass is iniially pulled down / f and released. (i) Wrie down he IVP for his sysem. Soluion: k = F/x = / =, so our IVP is / 64 x + x = 0; x(0) =, x (0) = 0. (ii) Solve he IVP. The general soluion will be x = A cos 8 + B sin 8. The iniial condiions give us x(0) = A = / and x (0) = 8B = 0. Then he soluion is x = cos 8. (iii) When does he mass reurn o he equilibrium posiion (i.e. x = 0). Soluion: x = 0 = π/6 for he firs ime.
(b) Now suppose he sysem has a damping consan of lb s/f. The mass is iniially pushed up / f and released wih a downward velociy of / f/s. (i) Wrie down he IVP for his sysem. Soluion: The damping adds a x erm, so 64 x + x + x = 0; x(0) =, x (0) =. (ii) Solve he IVP. Soluion: This is where he problem sars o be a pain, bu basically, he roos are r + 8r + 64 = 0 r = ( 8 ± ) 8 4 64 r, = 64 ± 8 63 From his poin les jus wrie down hings in he general form because i doesn make sense o carry all hose ridiculous number around. x = c e r + c e r. The firs iniial condiion gives us x(0) = c + c = / and he second iniial condiion gives us x (0) = r c + r c = /, hen we ge Moving on... c = r + (r r ), c = r + (r r ). () Given ha y = / is a soluion, please find anoher soluion o he ODE y + 3y + y = 0; > 0 Soluion: Le y = v()y y = v y + vy y = v y + v y + vy, Plugging ino he ODE gives us v y + v y + vy + 3v y + 3vy + vy = y v + ( y + 3y )v + 0 ( y + 3y + y )v = 0. Le u = v, y u = ( y + 3y )u u = ( + 3)u = u ln u = ln + c u = k y = k ln + c y = ln v = k ln + c du d u =
(3) Please solve he following IVP Soluion: soluion is y p acually is y + 4y = 3 sin ; y(0) =, y (0) =. The characerisic soluion is y c = A cos + A sin and our guess for he paricular =? B cos + B sin, bu look a ha, we have a repea, so our paricular soluion y p = B cos + B sin y p = B cos B sin + B sin + B cos y p = 4B sin 4B cos + 4B cos 4B sin. Plugging his ino he ODE gives 4B sin 4B cos + 4B cos 4B sin + 4B cos + 4B sin = 4B sin + 4B cos = 3 sin B = 0, B = 3 4. Then he general soluion is The derivaive of his is y = A cos + A sin 3 4 cos y(0) = A = y = A sin + A cos 3 4 cos + 3 sin y (0) = A 3 4 = A = 8. Then our soluion is y = cos 8 sin 3 cos. 4
Brief Summary of Chaper 3 Obviously incomplee so make sure you read he noes as well! Secion 3.: Exisence and Uniqueness and Wronskian (y + p(x)y + q(x)y = 0; y(0) = y 0, y (0) = Y 0 ) Exisence and Uniqueness: Pu ODE in sandard form and lis inervals for which he coefficiens and forcing funcion are coninuous, hen pick ou he inerval ha conains he iniial condiions. Wronskian: W (y, y ) = y y y y. Abel s Theorem: W (y, y ) = C exp ( x p(ξ)dξ ) Secion 3.3: Complex Roos: r = ξ ± iθ y = e ξx (A cos θx + B sin θx). Secion 3.4: Repeaed Roos and Reducion of Order Repeaed Roos: y = (c + c )e r. Reducion of Order: If y is a soluion, le y = v(x)y, plug i ino he ODE. The v(x) erm disappears and you re lef wih v and v erms. Le u = v, solve ha ODE, hen inegrae u o ge v. If you kep he consans of inegraion muliply v by y o ge your general soluion. Secion 3.5: Undeermined coefficiens: Find y c. Use f(x) o guess a y p and group like erms. If here are no repeas, ha s your y p. If here are repeas, ge rid of he repeas by muliplying hrough by x as many imes as needed. Secion 3.6: Variaion of Parameers: If we know y and y (someimes given, someimes found by solving he homogeneous equaion), hen y = y y f(x) d + y y f(x) W (y,y ) d. W (y,y ) Secion 3.7: Applicaions wihou forcing: Know how o se up and solve he IVPs. The ampliude in undamped oscillaory moion is he consan: R = A + B. For damped oscillaory moion R() = e ξ A + B. For undamped oscillaory moion he frequency and period are ω = k/m and T = π/ω. For damped oscillaory moion he quasi-frequency and quasi-period are θ (he imaginary par of he roo) and T d = π/θ. Secion 3.8: Applicaions wih forcing: Know your rig ideniies. The ransien soluion is he one ha goes o zero. The seady sae soluion is he one ha persiss. You can ge resonance in an undamped sysem if he frequency of your forcing funcion is he same as he frequency of your characerisic soluion because his causes a repea and he paricular soluion has o be muliplied by. You can ge resonance in an undamped sysem if damping is low enough and he frequency of your forcing funcion is close enough o he naural frequency, ω 0 (he frequency of he sysem in he absence of forcing and damping). Imporan ideniies: cos(a ± b) = cos a cos b sin a sin b and sin(a ± b) = sin a cos b ± cos a sin b. Typically, a = (ω 0 + ω) and b = (ω 0 ω) for his secion. Trig/Hyp ideniies ha may be useful overall: sin θ + cos θ = ; cosh x sinh x = cos(a ± b) = cos a cos b sin a sin b; sin(a ± b) = sin a cos b ± cos a sin b sin θ = ( cos θ); θ cos = ( + cos θ) You can basically derive any oher (seldom used) rig ideniy ha you may need from hese by eiher dividing hrough by a sin or a cos.