Methods in differential equations mixed exercise 7

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1 Methos in ifferential equations mie eercise 7 tan lnsec The integrating factor is e = e = sec Multiplying the equation by this factor gives: sec + ysectan= sec ( sec sec y = y sec= sec = tan+ c y= sin+ ccos 5 Rewrite the equation as + y= The integrating factor is e =e ln( =e ln( Multiplying the equation by this factor gives: 5 + y= ( ( ( y 5 = ( ( y 5 = ( ( y 5 = + c ( ( So y= 5 + c( =( Rewrite the equation as +y= The left-han sie is the erivative of the prouct y, so the equation is equivalent to (y= y= = +c So y= +c Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free.

2 ln The integrating factor is e = e = Multiplying the equation by this factor gives: + y= ( y = 5 y= = + c 5 c So y= The integrating factor is e =e Multiplying the equation by this factor gives: e y=e +e (e y=e e y= e = +c e So y= +ce Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free.

3 6 Rewrite the equation in the form The integrating factor is e ( = ( + y= ( ( an this simplifies as follows: = = ln( + ln( =ln( ln( =ln ( ( =ln ( So the integrating factor simplifies to ( Multiplying the equation by this factor gives: + y= ( ( ( y = ( ( y = c = + ( ( ( So y= + c( = + c 7 a Given that ay=keλ, the integrating factor is Multiplying the equation by this factor gives: a a λ a e ae y= ke e e a = e a ( e a e λ y = k e ( λ a a ( λ a ke ( λ a ye = ke = e + c λ a ( λ a a a ke e a ke a So y= + ce = + ce for λ a λ a λ a λ b When Q(=k n e a ( e e λ e e y = k = k e a a a λ n+ n+ a n k ye = k = + c n+ k a So y= e + ce n+ a Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free.

4 8 Rewrite the equation in the form + y tan =cos cot lnsin tan The integrating factor is e = e = e = sin Multiplying both sies by this factor gives: sin + y cos= cos sin= sin using a trigonometric ientity ( y sin = sin ysin= sin= cos+ c= + sin + c using cos= sin So y= sin+ Acosec efining A= c Note that without making the simplification using cos= sin, an alternative but correct answer is obtaine. ysin= sin= cos+c y= coscosec+ccosec 9 a The integrating factor is e tan =e lnsec = cos Multiplying both sies by this factor gives: cos +y sin cos =e y cos = e y cos =e +c So y=e cos+ccos b When =π,y= so e π c= c= e π So y=e cos (+e π cos Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free.

5 0 The integrating factor is e =e Multiplying the equation by this factor gives: e e y= e sin (e e y = sin y= e e sin ( The epression e sin can be integrate by using integration by parts twice e sin = e sin+ e cos using integration by parts with u=sin,v= e = e sin 9 e cos 9 e sin this time with u=cos,v= 9 e Simplifying this equation gives: 0 e sin 9 = e sin 9 e cos e sin= 0 e sin 0 e cos Using this epression in equation ( an aing in a constant gives: e y= e sin= 0 e sin 0 e cos+c So y= 0 sin 0 cos+ce Applying the conition that when =0,y=0 so 0 +c=0 c= 0 So the solution is y= 0 sin 0 cos+ 0 e a Separating the variables an integrating: =ysinh = sinh y lny=cosh+c wherecisa constant y=e cosh+c =e c e cosh y=ae cosh where Aisa constant (A=e c b When =0,y= so Ae= A=e So the solution is y=e e cosh =e cosh Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free. 5

6 Separating the variables an integrating: = ( y = y = rearranging the left-han sie so that it is easier to integrate ( ( + y y + y= ( ( + y y ln( y ln( y c + + = + + ln y y c = + + c + y e e = = A y + y = Ce y simplifying using the rules of logarithms taking the power of of each sie + y= Ce yce rearranging to fin an epression in y Ce y= Ce + When =0,y= so C C+ = C =C+ C= So the solution is y= 6e e + = (e e + The auiliary equation is m + m+ = 0 ± m= = ± i So the general solution is y= e Acos + Bsin If the auiliary equation has comple roots, the solution is of the form y=e p (Acosq+Bsinq The auiliary equation is m m+6=0 (m 6(m 6= 0 m=6 So the general solution is y=(a+be 6 If the auiliary equation has a repeate solution, the solution is of the form y=(a+be α Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free. 6

7 5 The auiliary equation is m m=0 m(m =0 m=0 or So the general solution is y=ae 0 +Be =A+Be 6 The auiliary equation is m +k =0 m=±ik The general solution is y=acosk+bsink When =0,y= so A= = kasink+kbcosk When = 0, = so kb= B= k Substituting values for A an B into the general solution gives the particular solution y= cosk+ sink k 7 The auiliary equation is m m+ 0= 0 ± 0 ± 6 m= = = ± i The general solution is y=e (Acos+Bsin When =0,y=0 so A=0 =e Bsin+e Bcos When = 0, = so B= B= Substituting values for A an B into the general solution gives the particular solution y=e sin If the auiliary equation has imaginary solutions, an the general solution has the form y= Acosω+ Bsin ω. 8 a y y= ke so = ke + ke an = ke + ke y Substituting into + y= e gives: ke +ke (ke +ke +ke =e 9ke =e k= 9 So a particular integral is e 9 Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free. 7

8 8 b Solving the corresponing homogeneous equation The auiliary equation is m m+=0 y + y= 0 ± 6 5 ± 6 m= = =±i So the complementary function is y=e (Acos+Bsin Using the particular integral from part a, the general solution is y=e (Acos+Bsin+ 9 e y 9 Solving the corresponing homogeneous equation The auiliary equation is m =0 (m+(m =0 m= or So the complementary function is y=ae +Be y = As the complementary function has an e term, try a particular integral in the form λe y So = λe + λe an = λe + λe y Substituting into y e = gives: λe +λe λe =e λe =e λ= So a particular integral is e The general solution is y=ae +Be +e 0 0 a Solving the corresponing homogeneous equation y +y=0 The auiliary equation is m m+=0 (m (m =0 m= So the complementary function is y=(a+be b As the complementary function has e an e terms, these cannot be terms in the particular integral. Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free. 8

9 0 c y=k e, so =k e +ke y =k e +ke +ke +ke =k e +8ke +ke Substituting into y y e + = gives: (k +8k+k 8k 8k+k e =e ke =e k= So a particular interval is e Using the complementary function from part a, the general solution is y=(a+b+ e Solving the corresponing homogeneous equation y t +y=0 The auiliary equation is m +=0 m=±i So the complementary function is y=acost+bsint The form of the particular integral is y=λcost+µsint y = λ sint+ µ cos t an = 9λ cost 9µ sint t t y Substituting into + y= 5cost gives: t 9λcost 9µsint+λcost+µsint=5cost 5λcost 5µsint=5cost λ= an µ=0 So a particular integral is cost The general solution is y=acost+bsint cost When t= 0, y= so A = A= = Asint+ Bcost+ sint t When t= 0, = so B= B= t Substituting values for A an B into the general solution gives the particular solution y=cost+sint cost Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free. 9

10 a Let y= λ+ µ + k e =µ+ke +ke y =ke +ke +ke =ke +ke Substituting into y y e + = + gives: ke +ke µ 6ke ke +λ+µ+ke =+e ke +µ+λ µ=+e Equating the coefficients of e : k= Equating the coefficients of : µ= µ= Equating constants: λ µ=0 λ= So a particular integral is ++e b Solving the corresponing homogeneous equation y +y=0 The auiliary equation is m m+=0 (m (m =0 m=or So the complementary function is y= Ae + Be Using the particular integral from part a, the general solution is y=ae +Be +++e =Ae +(B+e ++ a Solving the corresponing homogeneous equation 6 y +8 +5y=0 The auiliary equation is 6m + 8m+ 5= 0 8± m= = ± = ± i So the complementary function is e y= Acos + Bsin The particular integral has the form y=λ+µ, so =λ an y =0 Substituting into 6 y +8 +5y=5+ gives: 8λ+5λ+5µ=5+ Equating the coefficients of : 5λ=5 λ= Equating constant terms:8λ+5µ= µ= Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free. 0

11 a So a particular integral is + The general solution is e y= Acos + Bsin + + When = 0, y = so A + = A = 0 = e Acos + Bsin e Asin + Bcos + When = 0, y = so A+ B+ = As A = 0 B = Substituting values for A an B into the general solution gives the particular solution y=e Bsin ++ b As, e 0 so y + Therefore y + for large values of. Solving the corresponing homogeneous equation y 6y=0 The auiliary equation is m m 6=0 (m (m+=0 m=or So the complementary function is y=ae +Be The particular integral is of the form y=λsin+µcos = λ cos µ sin y = 9λ sin 9µ cos Substituting into y 6y sin cos = gives: 9λsin 9µcos λcos+µsin 6λsin 6µcos=sin cos (µ 5λsin (5µ+λcos=sin cos Equating the coefficients of sin : µ 5λ= ( Equating the coefficients of sin : 5µ+λ= ( Aing 5 equation ( to equation ( gives: 78µ= µ= 6 Substituting in equation ( gives: 5λ= 5λ= 5 = 6 Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free.

12 So a particular integral is (cos sin 6 The general solution is y= Ae + Be + (cos sin 6 When =0,y= so A+B+ 6 = A+B= 5 6 However as y remains finite for large values of, 5 A= 0 B= 6 Substituting values for A an B into the general solution gives the particular solution 5 y= e + (cos sin a Solving the corresponing homogeneous equation t +8 t +6=0 The auiliary equation is m +8m+6=0 (m+(m+=0 m= So the complementary function is y=(a+bte t The particular integral is of the form =λsint+µcost = λ cost µ sint t = 6λ sint 6µ cost t Substituting into t +8 +6=cost gives: t 6λ sint 6µ cost+ λ cost µ sint+ 6λ sint+ 6µ cost= cost µ sint+ λcost = cost λ = an µ = 0 So a particular integral is sint The general solution is =(A+Bte t + sint b When t=0,= so A= t t = ( A+ Bte + Be + cost t 8 When t= 0, = 0 so A+ B+ = 0 t 8 5 B= A = = Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free.

13 5 b Substituting values for A an B into the general solution from part a gives the particular solution = e t te t + sint c As t,e t 0 an te t 0,as e t ecreases faster than the linear factor. Challenge So for large values of t, the function behaves like = sint. It will oscillate like a sine wave with amplitue a Given that z= y, then y= z an z = z The equation (+ +y= y becomes z ( + z + z = z Multiply the equation by z + z = + + z + + gives ln( + The integrating factor is e = e = + (+ z +z= ( (+ z= (+ z= =+c z= +c + Asy=z, y= +c + b When = 0, y = so c= c= + So the particular solution is y = + π an perio. Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free.

14 Challenge u a Let = e, then u e u = u = u =eu = y u = u y + u = + y Fin y u in terms of an y, an y y show that u = + then substitute into the ifferential equation. So y + +y=ln transforms to y u + u +y=ln=u The auiliary equation is m +m+=0 (m+(m+=0 m= or So the complementary function is u y= Ae + Be Let the particular integral be of the form y=λu+µ, so u =λ an y u =0 Substituting into y u + +y=u gives: u λ+λu+µ=u u Equating the coefficients of u: λ= λ= Equating constants: λ + µ = 0 µ = µ = So a particular integral is u u u The general solution is y= Ae + Be + u But =e u u=ln an e u = =, e u = = So the general solution of the original equation is y= A + B + ln Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free.

15 Challenge b When =,y= so A+B = A+B= 7 ( A B = + When =, = so A B+ = A+B= ( Subtracting equation ( from equation ( gives B= an so from equation ( A = + = Substituting values for A an B into the general solution from part a gives the particular solution y= 9 + ln Pearson Eucation Lt 08. Copying permitte for purchasing institution only. This material is not copyright free. 5

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