Methods in differential equations 7B

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1 Methods in differential equations 7B a The auxiliary equation is m +5m+6= (m+)(m+)= m= or y=ae x +Be x d y dy The auxiliary equation of a b cy + + = is am + bm + c=. If αandβ are real roots of this quadratic then e αx βx y= A is the general solution of the differential equation. m 8m+= (m 6)(m )= m=or 6 y=ae x +Be 6x c The auxiliary equation is m +m 5= (m+5)(m )= m= 5or y=ae 5x +Be x m m 8= (m 7)(m+4)= m=7 or 4 y=ae 7x +Be 4x e The auxiliary equation is m +5m= m(m+5)= m= or 5 The auxiliary equation has two real roots, but one of them is zero. As e x =, the general solution is y=ae x +Be 5x =A+Be 5x m +7m+= (m+)(m+)= m= or x x Pearson Education Ltd 8. Copying permitted for purchasing institution only. This material is not copyright free.

2 g The auxiliary equation is 4m 7m = (4m+)(m )= m= 4 or x 4 x h The auxiliary equation is 5m 7m = (5m+)(m )= m= 5 or x x 5 a The auxiliary equation is m +m+5= (m+5)(m+5)= m= 5 y=(a+bx)e 5x d y dy The auxiliary equation of a b cy + + = is am + bm + c=. If this equation has one repeated real rootα then the general solution of the differential equation is y=(a+bx)e αx m 8m+8= (m 9)(m 9)= m=9 y=(a+bx)e 9x c The auxiliary equation is m +m+= (m+)(m+)= m= y=(a+bx)e x m 8m+6= (m 4)(m 4)= m=4 y=(a+bx)e 4x Pearson Education Ltd 8. Copying permitted for purchasing institution only. This material is not copyright free.

3 e The auxiliary equation is 6m +8m+= (4m+)(4m+)= m= 4 4 y= ( A+ 4m 4m+= (m )(m )= m= y = ( A+ g The auxiliary equation is 4m +m+5= (m+5)(m+5)= m= 5 5 y= ( A+ h The auxiliary equation is m + m+= (m+ )(m+ )= m= y=(a+bx)e x a The auxiliary equation is m +5= m=±5i The general solution is y=acos5x+bsin5x m +8= m=±9i The general solution is y= Acos9x+ Bsin9 x. If the auxiliary equation has purely imaginary roots, the general solution has the form y=acosωx+bsinωx, where A and B are constants and iω is the solution of the auxiliary equation. Pearson Education Ltd 8. Copying permitted for purchasing institution only. This material is not copyright free.

4 c The auxiliary equation is m += m=±i The general solution is y=acosx+bsinx 9m +6= m = 6 9 m=± 4 i The general solution is 4 4 y= Acos x+ Bsin x e The auxiliary equation is = 8± m= = 4± 4= 4± i using the quadratic formula The general solution is y=e 4x (Acosx+Bsinx) 4 + 5= 4± 6 m= = ± 4= ± i The general solution is y=e x (Acosx+Bsinx) g The auxiliary equation is + + 9= ± 4 46 ± 6 m= = = ± i The general solution is x e ( Acosx Bsin x) y= + If the auxiliary equation has complex roots, the general solution has the form y=e px (Acosqx+Bsinqx),where A and B are constants and p± iq are solutions of the auxiliary equation. h The auxiliary equation is m + m+= m= ± = The general solution is ± 9 = ±i x y= e Acos x+ Bsin x Pearson Education Ltd 8. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 4 a The auxiliary equation is m +4m+49= (m+7)(m+7)= m= 7 y=(a+bx)e 7x m +m = (m+4)(m )= m= 4 or y=ae 4x +Be x c The auxiliary equation is = 4± 6 5 4± 6 m= = = ± i The general solution is y=e x (Acosx+Bsinx) 6m 4m+9= (4m )(4m )= m= 4 4 y= ( A+ e The auxiliary equation is 9m 6m+ 5= 6± ± 6 8 6± 44 6± i ± i m= = = = = x y e = Acos x+ Bsin x 6m m = (m )(m+)= m= or x x Pearson Education Ltd 8. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 5 a The auxiliary equation is m +km+9= m= k± 4k 6 = k± k 9 using the quadratic formula i If k >, the auxiliary equation has two real solutions So the differential equation has the general solution x=ae ( k+ k 9)t +Be ( k k 9)t ii If k <, then the auxiliary equation has two complex conjugate roots, k±i 9 k (( ) ) t x=e kt Acos 9 k (( ) ) t +Bsin 9 k iii If k =, then the solution of the auxiliary equation is one repeated root k, So the differential equation has general solutionx=(a+bt)e kt ; b i The general solution can be found simply substituting k = in the equation found in part a ii ( ) x=e t Acos 5t+Bsin 5t ii When t, e t, while the trigonometric functions take values that are bounded, so x. 6 If α is the only root of the equation, then: () aα +bα+c= by definition () α = b a from the quadratic formula, as equal roots b =4ac Let y=(a+bx)e αx Then dy =Beαx +(A+Bx)αe αx And d y =Bαeαx +Bαe αx +(A+Bx)α e αx = Bαe αx +(A+Bx)α e αx Substituting these results into the differential equation gives: a d y +bdy +c=a(bαeαx +(A+Bx)α e αx )+b(be αx +(A+Bx)αe αx )+c(a+bx)e αx =Be αx (aα+b)+(a+bx)e αx (aα +bα+c) But from equation () aα +bα+c= and from equation () α = b a aα+b= Hence a d y +bdy +c=, and so y=(a+bx)eαx is a solution to this equation. Pearson Education Ltd 8. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 7 Lety=Af(x)+Bg(x) Then dy =Adf(x) +B dg(x) And d y f(x) =Ad +B d g(x) Then substituting these results into the differential equation gives: a d y +bdy +c=a f(x) Ad +B d g(x) +b Adf(x) +B dg(x) =A a d f(x) +b df(x) +cf(x) +B g(x) ad +c( Af(x)+Bg(x) ) +b dg(x) +cg(x) As y=f(x) and y=g(x) are solutions of the differential equation, it follows that a d f(x) +b df(x) +cf(x)= and a d g(x) +b dg(x) Therefore A a d f(x) +b df(x) +cf(x) +B g(x) ad So y=af(x)+bg(x) is a solution. Challenge +cg(x)= +b dg(x) +cg(x) = If a real-valued quadratic equation has complex roots p±qi then by Euler s formula, e p+iq =e p e iq =e p (cosq+isinq) e p iq =e p e iq =e p (cos( q)+isin( q)=e p (cosq isinq) So substituting for α =p+iq and β =p iq into Ae αx +Be βx gives: Ae αx +Be βx =Ae px (cosqx+isinqx)+be px (cosqx isinqx) =e px ((A+B)cosqx+(A B)isinqx) Choose A and B such that they are complex conjugates, i.e. there are real numbers λ and µ where A=λ+µi and B=λ µi So A+B=λ and (A B)i=(ui)i=ui = u Hence Ae αx +Be βx =e px (λcosqx µsinqx)=e px (Ccosqx+Dsinqx), where C and D are real constants, C=λ and D= µ Pearson Education Ltd 8. Copying permitted for purchasing institution only. This material is not copyright free. 7