Chapter 16



CHAPTER 15 SOLUTIONS TO EXERCISES IN ORDINARY DIFFERENTIAL EQUATIONS

Exercise on 15.1

Solve the more general equation (15.1) in the case when the initial number of bacteria is n0 = 4. If time is measured in seconds, and after 5 seconds it is found that the number of bacteria is 10, what is the value of k?

Solution

We have

= kn

The general solution is

n = A ekt

When t = 0, n(0) = A = n0 and so

n = 4 ekt

Now when t = 5, n = 10 and so

10 = 4 e5k

So

e5k =

Taking 'antilogs', or exponentiating we have

5k = ln

and therefore

k = ln

Exercises on 15.2

1. State the order of the following differential equations. Which are nonlinear?

i) = ex + 1 ii) – 9y = 0

iii) y + cos x = 0 iv) + 2y2 = 1

v) – 4 + 3y = 3x + 2

Solution

i) The highest derivative in = ex + 1 is clearly one, so it is a first order equation. Also, the highest power of y or its derivative is also one, so this equation is linear.

ii) The highest derivative in – 9y = 0 is two, so this is a second order equation - it is also linear since no powers of y or its derivatives occur higher than one.

iii) y + cos x = 0 is second order, but is nonlinear because of the y multiplying the second derivative.

iv) + 2y2 = 1 is third order, but again nonlinear both because of the multiplying the second derivative and the y2

v) – 4 + 3y = 3x + 2 is second order and linear

2. Verify that the following functions are each solutions of one of the equations in Q1, and match the solution to its equation.

a) 2e3x b) ex + x + 2

c) e3x + x + 2

Solution

On the face of it we need to try each function in each equation, and if you want plenty of practice in differentiation, by all means do! However, with a little common sense we can save ourselves a lot of work. For example none of these functions contains a trig function, and this would rule out equation iii). Also, we know that when we differentiate an exponential, we always get back a multiple of the same exponential, and so equation iv) could not eliminate the exponentials in the proposed solutions, ruling out this equation. So we are left with equations i), ii), iii) to try.

a) 2e3x could not be a solution of i) because differentiating it could not produce the 1 - and similarly for equation v). We therefore only need to check that it is a solution of ii). We have, with y = 2e3x

– 9y ∫ – 9

= 2 – 18 e3x

2 – 18 e3x = 18e3x – 18 e3x = 0

So y = 2e3x is a solution of the equation ii) – 9y = 0

b) y = ex + x + 2 is most likely to be a solution of i) because c) would not give the ex in i). This is easily confirmed:

= = ex + 1

as required. So y = ex + x + 2 is a solution of i) = ex + 1

c) y = e3x + x + 2 must now be a solution of the only remaining equation v), which we now check

– 4 + 3 (e3x + x + 2)

= 9 e3x – 12 e3x – 4 + 3 e3x + 3 x + 6

= 3 x + 2 on simplification

So y = e3x + x + 2 is a solution of v) – 4 + 3y = 3x + 2

3. Find the general solution of equation 1 i) and the particular solution that satisfies y(0) = 1.

Solution

1 i), = ex + 1, is a particularly simple equation that can be 'solved' by direct integration. We obtain

y = dx + C = ex + x + C

where C is an arbitary constant. This is the most general solution - it has been obtained by direct integration, and we cannot find any more solutions. So the general solution is

y = ex + x + C

If y(0) = 1 then we have

y(0) = 1 = e0 + 0 + C = 1 + C

so C= 0 and the particular solution in this case is

y = ex + x

Exercise on 15.3

Solve the differential equations

i) y´= sin x ii) y´= y2

iii) y´ = x2y iv) xy´ = 2x + y

In i), ii), iii) give the particular solutions satisfying the condition y(0) = 1. In iv) give the solution satisfying y(1) = 0.

Solution

i) y´= sin x can be solved by direct integration. We have

y = sin x dx + C

= – cos x + C

which is the required solution, as you should check by substituting back into the DE. If y(0) = 1 then we have

y(0) = 1 = – cos 0 + C = – 1 + C

and so C = 2 and the particular solution is

y = 2 – cos x

ii) y´= = y2 is a variables separable equation and we can write

" = dx "

So, integrating both sides with respect to their respective variables we have

= – = dx + C = x + C

So, rearranging we have

y =

Now when y(0) = 1, we have

y(0) = 1 =

or C = 1, giving the particular solution

y =

iii) y´ = x2y is a separable equation also and we can write

" = x2 dx "

So, integrating both sides with respect to their respective variables we have

= ln y = x2 dx + C = + C

So, exponentiating we get

y = exp= eC exp

We can replace eC by a new arbitrary constant A if we wish, or simply rename C (since it is arbitrary anyway) to write the general solution as

y = C exp

Since y(0) = 1 we find C = 1 and so the particular solution is

y = exp

iv) The equation xy´ = 2x + y can be rewritten

y´ = = 2 +

and is therefore homogeneous, and can be solved by substituting y = xv:

y´ = xv´ + v = 2 + v

and so

xv´ = 2 = x

This equation is separable and gives

2= 2 ln x = dv = v + C

So

v = 2 ln x + C =

The GS is then

y = 2x ln x + Cx

Now y(1) = 0 gives 0 = 0 + C (ln 1 = 0), and so C = 0 and the particular solution is

y = 2x ln x

Exercise on 15.4

Find integrating factors for the following equations and hence obtain the general solution

i) xy´+ y = x ii) xy´– 2y = x3 + 2

Can you dispense with the integrating factor, by finding a derivative of a product?

Solution

i) First convert xy´+ y = x to standard form

y´ + y = 1

Now multiply through by an integrating factor I:

I y´ + y = I

Put

=

Solve this separable equation for I

=

giving

ln I = ln x or I = x

(remember we don’t need an arbitrary constant here)

Multiply through by the integrating factor to retrieve the original equation, the LHS of which we now know can be written as the derivative of a product

x + y = = = x

Now integrate (now we bring in the arbitrary constant)

xy = + C

or

y = +

Now the very fact that multiplying by the integrating factor returned us to the original 'non-standard' form of the equation leads us to suspect that we really didn't need to use the IF method at all. And indeed, if your differentiation is up to scratch you should notice that

xy´+ y = x + y = = x

which can be integrated directly.

ii) Converting xy´– 2y = x3 + 2 to standard form gives

y´– y = x2 +

The equation for the integrating factor then becomes

= – I

Solving this for I gives

= –

or

ln I = – 2 lnx = ln x– 2

So

I = x– 2 =

Multiply the DE through by I to get

– y = = 1 +

Now we can integrate through to get

y = x – + C

or

y = x3 + C x2 – 1

In this case it is not so easy to spot the total derivative that the original equation represented but the – 2 in xy´– 2y gives a hint that we have differentiated a x–2, so we turn the coefficent of dy/dx into this by dividing by x–3 to get – y which we then (hopefully) recognize as as we found using the integrating factor.

Exercises on 15.5

1. Solve the following initial value problems

i) y´´– y´– 6y = 0 y(0) = 1 y´(0) = 0

ii) 2y´´+ y´– 10y = 0 y(0) = 0 y´(0) = 1

Solution

i) y´´– y´– 6y = 0 y(0) = 1 y´(0) = 0

Put y = eλx to obtain the auxiliary equation

λ2 – λ – 6 = (λ – 3)(λ + 2) = 0

from which we obtain λ = – 2, 3 and so the GS is

y = Ae3x + Be–2x

Applying the initial conditions y(0) = 1, y´(0) = 0 gives

A + B = 1

3A – 2B = 0

which yield

A = and B =

and so the particular solution is

y = e3x + e–2x

ii) 2y´´+ y´– 10y = 0 y(0) = 0 y´(0) = 1

Put y = eλx to obtain the auxiliary equation

2λ2 + λ – 10 = (λ – 2)(2λ + 5) = 0

ie

λ = 2, –

and so the GS is

y = Ae2x + Be–5x/2

Applying the initial conditions y(0) = 0, y´(0) = 1 gives

A + B = 0

2A – B = 1

or

4A – 5B = 2

which yield

A = and B = –

and so the particular solution is

y = e2x – e–5x/2

2. Solve the following boundary value problems

i) y´´+ 4y´+ 13y = 0 y(0) = 0 y= 1

ii) y´´– 4y´+ 4y = 0 y(0) = 0 y(1) = 1

Solution

i) y´´+ 4y´+ 13y = 0 y(0) = 0 y= 1

The auxiliary equation is

λ2 + 4λ + 13 = 0

The solution of this quadratic is found to be

λ = – 2 ± 3j

So the GS is

y = e–2x(A cos 3x + B sin 3x)

in 'real' form. The BCs give

y(0) = A = 0

y= e–π B sin = – e–π B = 1

so

B = – eπ

and the particular solution is

y = e–2x( – eπ sin 3x)

= – eπ – 2x sin 3x

ii) y´´– 4y´+ 4y = 0 y(0) = 0 y(1) = 1

The AE is

λ2 – 4λ + 4 = (λ – 2)2 = 0

So the GS is

y = (Ax + B) e2x

The BCs give

y(0) = B = 0

y(1) = Ae2 = 1, so A = e–2

The required solution is therefore

y = e–2 x e2x = x e2(x – 1)

Exercises on 15.6

1. Find the solutions to each of the following second order equations, with the specified conditions. Remember to apply the conditions to the full solution – CF + PI.

i) y´´+ 4y´+ 3y = 2ex y(0) = 0 y´(0) = 1

ii) y´´+ 4y = x + 1 y(0) = 0 y() =

iii) y´´+ y = sin 2x y(0) = 0 y′(0) = 0

Solution

i) y´´+ 4y´+ 3y = 2ex y(0) = 0 y´(0) = 1

We first find the complementary function by finding the GS of the homogeneous equation:

y´´+ 4y´+ 3y = 0

The AE is

λ2 + 4λ + 3 = (λ + 1)(λ + 3) = 0

giving

λ = – 1 and – 3

So the CF is

yc = A e –x + B e –3x

To find a particular integral we look for any solution of the inhomogeneous equation

y´´+ 4y´+ 3y = 2ex

We note that the RHS 2ex is not contained in the CF, so we are safe in trying a solution of the form

yp = Lex

Substituting into the equation gives

(L + 4L + 3L) ex= 2ex

From which

8L = 2 or L =

So a PI is

yp = ex

The GS is therefore

y = yc + yp = A e –x + B e –3x + ex

We now apply the ICs

y(0) = A + B + = 0

y´(0) = – A – 3B + = 1

Or

A + B = –

A + 3B = –

giving

A = 0 and B = –

Hence the solution required is

y = – e –3x + ex =

ii) y´´+ 4y = x + 1 y(0) = 0 y() =

The CF is the general solution of y´´+ 4y = 0, ie

yc = A cos 2x + B sin 2x

For the PI we try a linear function, like the RHS

yp = Lx + M

Substituting into the equation gives

4(Lx + M) = x + 1

from which

4L = 1 and 4M = 1, or L = M =

A PI is therefore

yp = x +

The GS of the inhomogeneous equation is thus

y = yc + yp = A cos 2x + B sin 2x + x +

Applying the BCs gives

y(0) = A + = 0 or A = –

and

y= A cos + B sin + + =

ie

B + = 0 or B = –

The required solution is therefore

y = – cos 2x – sin 2x + x +

iii) y´´+ y = sin 2x y(0) = 0 y′(0) = 0

The CF is

yc = A cos x + B sin x

For the PI we can on this occasion use yp = L sin 2x, since there is no y´ term in the differential equation, and substituting into the equation gives

– 4L sin 2x + L sin 2x = sin 2x

from which

L = –

and so the PI is

yp = – sin 2x

and the GS is

y = yc + yp = A cos x + B sin x – sin 2x

The ICs give

y(0) = A = 0

y′(0) = B cos 0 – cos 0 = B – = 0 so B =

The required solution is therefore

y = sin x – sin 2x

2. Solve the initial value problem

y´´– 4y´ + 3y = 3x y(0) = 0 y´(0) = 0

Solution

The CF is the general solution of

y´´– 4y´ + 3y = 0

The AE is

λ2 – 4λ + 3 = (λ – 1)(λ – 3) = 0

So λ = 1 and λ = 3 and the CF is

yc = A e3x + B ex

For the PI we try

yp = Lx + M

in the equation to get

– 4L + 3Lx + 3M = 3x

From this we get

L = 1 and 3M – 4 = 0, or M =

and a PI is

yp = x +

The GS is therefore

y = yc + yp = A e3x + B ex + x +

Applying the ICs gives

y(0) = A + B + = 0

and

y´(0) = 3A + B + 1 = 0

giving

A = and B = –

So the required solution is

y = e3x – ex + x +

REINFORCEMENT EXERCISES

1. Radioactive material decays at a rate proportional to the amount present: construct and solve a mathematical model giving the amount of material remaining after a given time.

Solution

Let the mass at time t be m. Then 'radioactive material decays at a rate proportional to the amount present' can be expressed mathematically as

∝ m or = – λm

where λ is a positive constant - the negative sign denoting a decaying process. By now you should be able to state the GS immediately as

m = m0 e– λt

where m0 is an arbitrary constant.

2. Solve the following differential equations subject to the conditions given:

i) y´= x y(0) = 1

ii) y´= cos x y(() = 0

iii) xy´= x2 + 1 y(1) = 0

iv) y´´= 4 y(0) = 1 y´(0) = 2

v) y´´= x2 – 1 y(0) = 0 y (2) = 1

vi) y´´= cos x y(0) = 0 y(() = 1

vii) y´= 3y2 y(0) = 1

viii) y´= sec y y(0) = (

Solution

i) y´= x y(0) = 1

This is a case of direct integration

y = x dx + C = + C

The IC gives

y(0) = C = 1

so the solution required is

y = + 1

ii) y´= cos x y(() = 0

Again direct integration

y = cos x dx + C = sin x + C

The IC gives (note that an IC does not necessarily have to be specified at x = 0 !)

y(π) = C = 0

So the required solution is

y = sin x

iii) xy´= x2 + 1 y(1) = 0

After dividing through by x this again yields direct integration

y´= = x +

and integration gives

y = + ln x + C

The IC gives

y(1) = + ln 1 + C = 0 or C = –

and the particular solution is

y = + ln x –

iv) y´´= 4 y(0) = 1 y´(0) = 2

This time we have to integrate twice. After one integration we get

y´= 4x + C

and another gives

y = 2x2 + Cx + D

The ICs give

y´(0) = C = 2

y(0) = D = 1

So the solution is

y = 2x2 + 2x + 1

v) y´´= x2 – 1 y(0) = 0 y (2) = 1

Again integrating twice gives

y = – + Cx + D

The BCs give

y(0) = D = 0

y(2) = – + 2C = 1, so C =

and the solution is finally

y = – + x

vi) y´´= cos x y(0) = 0 y(() = 1

Two integrations give

y = – cos x + Cx + D

The BCs give

y(0) = – 1 + D = 0 so D = 1

y(() = – cos ( + (C + 1 = 1

from which

C = =

and the solution is

y = – cos x – x + 1

vii) y´= 3y2 y(0) = 1

By turning both sides upside down we obtain a direct integration with respect to y

y´= 3y2 =

so

=

Integrating with respect to y gives

x = – + C

The IC y(0) = 1 now gives

0 = – + C, so C =

and the solution is therefore

x = – + =

rearranging:

3x – 1 = –

so

y =

viii) y´= sec y y(0) = (

Again turning both sides upside down we have

= sec y =

so

= cos y

and

x = sin y + C

The IC y(0) = ( gives

0 = sin ( + C , so C = 0

The solution is therefore

x = sin y

or

y = sin–1 x

3. Find the general solution of the differential equations y´= f(x, y) where f(x, y) is given by

i) xy2 ii) iii) x sec y

iv) ex+y v) 10 – 2y vi) e2x – 3y

vii) viii) y lnx ix) (x – y)/x

x) y(x + 2y)/[x(2x + y)] xi) 3(y2 – 3y + 2)

Solution

i) The equation y´= xy2 is separable, and we can write

= – = x dx + C = + C

Solving for y we thus get the GS

y = – = – on redefinition of C

ii) = is also separable, giving

= ln y = = ln x + C = ln(Cx)

and so

y = Cx

iii) = x sec y = is again separable, and gives

cos y dy = x dx

So

sin y = + C

and the GS is

y = sin–1

iv) = ex+y = ex ey is separable

e–y dy = ex dx + C

So

– e– y = ex + C

e–y = C – ex

– y = ln(C – ex)

Giving the GS

y = ln

NOTE: Strictly, all expressions in which, as above, we take a log of a function, we should include modulus signs, since we can only take the log of a positive quantity. However, to avoid cluttering up expressions, it is usual to omit this, it being taken for granted.

v) = 10 – 2y is separable – and also linear. The easiest way to tackle it is clearly by separation of variables however.

= dx + C

So, integrating

– ln (10 – 2y) = x + C

or

ln(10 – 2y) = C – 2x

Exponentiating gives

10 – 2y = eC – 2x

from which we find, on redefining C

y =

vi) = e2x – 3y = e2x e– 3y separates to give

e3y dy = e2x dx + C

Integrating

e3y = + C

Rearranging

e3y = + C

Taking logs

3y = ln

Finally giving the solution

y = ln

vii) =

5 – sin y dy = 5y + cos y = 2x dx + C = x2 + C

So the solution is

5y + cos y = x2 + C

Note that in this case we cannot solve explicitly for y, so we leave the solution as it is, in implicit form

viii) = y lnx yields, on using the standard integral for the natural log

= ln y = ln x dx + C = x ln x – x + C

So

y = ex ln x – x + C = Cex ln x – x (x > 0)

= C xx e– x

ix) = (x – y)/x = = 1 – is homogeneous, so we put v = or y = xv to get

= v + x = 1 – v

So

x= 1 – 2v

Therefore, separating the variables,

= + C

So integrating both sides

ln(1 – 2v) = ln x + C = ln (Cx)

Taking anti-logs

1 – 2v = Cx

Or

v = =

So finally

y =

x) = y(x + 2y)/[x(2x + y)] = is again homogeneous and substituting y = xv gives

= v + x =

and so

x = – v =

Separating the variables gives

= + C

We can integrate the left-hand side using partial fractions

dv = ln (Cx)

Therefore

3 ln(v – 1) – 2 ln v = ln = ln Cx

Removing the logs gives

= Cx

3= C v2 x = C x = C

which simplifies finally to

(y – x)3 = Cx2 y2

xi) = 3(y2 – 3y + 2) is separable:

= 3dx + C = 3 x + C

The LHS can be integrated using partial fractions

= dy

= ln (y – 2) – ln (y – 1) = ln = 3x + C

So, removing logs

= Ce3x

4. Solve the equations

i) y´= e2x – y ii) y´+ 2y = 3ex iii) y´+ xy = x3

iv) xy´= 3y – 2x v) (x2 – 1)y´+ 2y = 0

vi) (x – 1)y´= 3x2 – y vii) xy´– 2y = x3e–2x

Solution

All of these equations are linear. We will only illustrate the explicit determination of the integrating factor in the first example, leaving you to confirm the remaining IFs as exercises.

i) y´= = e2x – y rearranges into the standard form

+ y = e2x

Now multiply through by an integrating factor I:

I + Iy = Ie2x

Put

= I

Solve this separable equation for I

= dx

giving

ln I = x or I = ex for the integrating factor

Multiply through by the integrating factor to obtain the derivative of a product

ex + exy = = = exe2x = e3x

Integrating this directly now gives

ex y = e3x dx + C = e3x + C

So, dividing through by ex

y = Ce–x + e2x

ii) The IF for y´+ 2y = 3ex is e2x and multiplying through by it gives

e2x + 2e2x y = = 3 e3x

Therefore, integrating:

e2x y = e3x + C

or

y = ex + C e–2x

iii) For y´+ xy = x3 the IF is exp(x2/2), so

exp(x2/2) + x y exp(x2/2) = = x3 exp(x2/2)

Integrating both sides gives

exp(x2/2) y = x3 exp(x2/2) dx + C

To integrate the RHS put u = , so du = xdx and

x3 exp(x2/2) dx = x2 exp(x2/2) (xdx)

= 2u exp u du = 2

on integrating by parts

= 2(ueu – eu) = 2eu(u – 1) = exp(x2/2) (x2 – 2)

So

exp(x2/2) y = exp(x2/2) (x2 – 2) + C

or

y = x2 – 2 + Cexp(– x2/2)

iv) xy´= 3y – 2x rearranges to the standard form

y´ – y = – 2

The IF is and so

– y = = –

Therefore

= + C

or

y = x + C x3

v) (x2 – 1)y´+ 2y = 0 is actually separable, as well as linear, and in fact the LHS is immediately recognizable as the derivative of a product

(x2 – 1)y´+ 2y = = 0

which integrates to

(x2 – 1)y = C

or

y =

vi) (x – 1)y´= 3x2 – y becomes

(x – 1)y´ + y = = 3x2

Integrating gives

(x – 1) y = x3 + C

or

y =

vii) xy´– 2y = x3e–2x rearranges to

y´– y = x2e–2x

The IF is found to be giving

y´– y = e–2x =

So

y = – e–2x + C

and hence the GS is

y = Cx2 – e–2x

5. Solve the following initial value problems

i) y´– 3y = e5x y(0) = 0

ii) xy´– 2y = x2 y(1) = 2

iii) xy´+ 2y = x2 y(1) = 0

iv) xy´+ 3y = y(π) = 0

Solution

i) y´– 3y = e5x y(0) = 0

The IF is e–3x:

e–3x y´– 3e–3x y = e2x =

So

e–3x y = e2x + C

or

y = e5x + C e3x

Now applying the ICs gives

y(0) = 0 = + C

so C = – and therefore the solution is:

y = = e3x

ii) xy´– 2y = x2 y(1) = 2

The standard form is

y´– y = x

The IF is :

y´– y = =

Therefore

y = ln x + C

or

y = x2 ln x + C x2

The ICs now give y(1) = C = 2, so finally the solution becomes

y = x2 (ln x + 2)

iii) xy´+ 2y = x2 y(1) = 0

An ‘obvious’ multiplication by x gives

x2 y´+ 2xy = = x3

So, integrating

x2 y = + C

or

y = +

The IC y(1) = 0 now gives + C = 0

So C = ( and the solution is

y = (

iv) xy´+ 3y = y(π) = 0

Multiplying through by x2 gives (The 3 is the hint here – it comes from differentiating x3)

x3 y´+ 3x2 y = = sin x

So

x3 y = – cos x + C

y = – +

Now y(π) = 0 gives – (– 1) + C = 0 or C = – 1, so the required solution is

y = – – = –

6. Solve the following second order equations

i) y´´+ y´– 2y = 0 ii) y´´– 4y´+ 4y = 0

iii) y´´+ 4y´+ 5y = 0 iv) y´´+ 4y = 0

v) y´´– 9y = 0 vi) y´´+ y = 0

Solution

i) y´´+ y´– 2y = 0

Substituting y = eλx gives the AE

λ2 + λ – 2 = (λ – 1)(λ + 2) = 0

So

λ = 1, – 2

and therefore the GS is

y = Aex + Be–2x

ii) y´´– 4y´+ 4y = 0

Substituting y = eλx gives the auxiliary equation

λ2 – 4λ + 4 = (λ – 2)2 = 0

So

λ = 2 (twice)

and therefore the GS is

y = (Ax + B)e2x

iii) y´´+ 4y´+ 5y = 0

Substituting y = eλx gives the AE

λ2 + 4λ + 5 = 0

So

λ = = = – 2 ± j

and therefore the GS is

y = e–2x (Acos x + B sin x)

iv) y´´+ 4y = 0

Substituting y = eλx gives the AE

λ2 + 4 = 0

So

λ = ± 2j

and therefore the GS is

y = Acos 2x + B sin 2x

v) y´´– 9y = 0

The AE is

λ2 – 9 = 0

So

λ = ± 3

and therefore the GS is

y = Ae3x + Be–3x

vi) y´´+ y = 0

The AE is

λ2 + 1 = 0

So

λ = ± j

and therefore the GS is

y = Acos x + B sin x

7. Obtain the general solution of the inhomogeneous equations formed by adding the following right hand sides to each of the equations of Q6.

a) 2 b) x + 1 c) e–2x

d) 2 sin x

Solution

a) i) y´´+ y´– 2y = 2

The CF, ie the GS of the homogeneous equation, is from Q 6 i)

yc = Aex + Be–2x

For a particular integral we try a solution y = L and obtain on substitution in the equation

– 2L = 2 or L = – 1

So a particular solution is yp = – 1 and the GS of the equation a) i) is therefore

y = yc + yp = Aex + Be–2x – 1

a) ii) y´´– 4y´+ 4y = 2

From 6 ii) the GS to y´´– 4y´+ 4y = 0 gives the CF as

yc = (Ax + B)e2x

Again using y = L as a trial solution for the PI gives for a PI:

yp = =

So the GS is

y = yc + yp = (Ax + B)e2x +

a) iii) y´´+ 4y´+ 5y = 2

From 6 iii) the GS to y´´+4y´+ 5y = 0 gives the CF as

yc = e–2x (Acos x + B sin x)

A PI is found to be

yp =

So the GS is

y = yc + yp = e–2x (Acos x + B sin x) +

a) iv) y´´+ 4y = 2

From 6 iv) the GS to y´´+ 4 = 0 gives the CF as

yc = Acos 2x + B sin 2x

A PI is found to be

yp =

So the GS is

y = yc + yp = Acos 2x + B sin 2x +

a) v) y´´– 9y = 2

The CF is

yc = Ae3x + Be–3x

A PI is

yp = –

So the GS is

y = yc + yp = Ae3x + Be–3x –

a) vi) y´´+ y = 2

The CF is

yc = Acos x + Bsin x

A PI is

yp = 2

So the GS is

y = yc + yp = Acos x + Bsin x + 2

b) i) y´´+ y´– 2y = x + 1

yc = Aex + Be–2x

For a particular integral we try a solution yp = Lx + M and obtain on substitution in the equation

yp'' + yp' –2 yp = L –2(Lx + M) = – 2Lx + L – 2M ( x + 1

so

– 2L = 1 or L = –

and

L – 2M = 1, so M = = –

Therefore

yp = – x –

and the GS of the equation is

y = yc + yp = Aex + Be–2x – x –

b) ii) y´´– 4y´+ 4y = x + 1

yc = (Ax + B)e2x

Again using y = Lx + M as a trial solution gives for a PI:

yp = x +

So the GS is

y = yc + yp = (Ax + B)e2x + x +

b) iii) y´´+ 4y´+ 5y = x + 1

yc = e–2x (Acos x + B sin x)

With the trial yp = Lx + M we have

y´´+ 4y´+ 5y = 4L + 5(Lx + M) = 5Lx + 4l + 5M ( x + 1

Equating coefficients gives

5L =1

5M = 1 ( 4L

from which we find

L = and M =

yp = x +

So the GS is

y = yc + yp = e–2x (Acos x + B sin x) + x +

b) iv) y´´+ 4y = x + 1

yc = Acos 2x + B sin 2x

A PI is found to be

yp = x +

So the GS is

y = yc + yp = Acos 2x + B sin 2x + x +

b) v) y´´– 9y = x + 1

The CF is

yc = Ae3x + Be–3x

A PI is

yp = – x –

So the GS is

y = yc + yp = Ae3x + Be–3x – x –

b) vi) y´´+ y = x + 1

yc = Acos x + Bsin x

A PI is

yp = x + 1

So the GS is

y = yc + yp = Acos x + Bsin x + x + 1

c) i) y´´+ y´– 2y = e–2x

yc = Aex + Be–2x

For a particular integral the trial yp = Le–2x suggested by the RHS will not work because this is already contained in the CF. We therefore try yp = Lxe–2x and obtain on substitution in the equation:

yp'= Le–2x – 2Lxe–2x

yp'' = –4Le–2x + 4Lxe–2x

so

yp'' + yp' –2 yp = –4Le–2x + 4Lxe–2x + Le–2x – 2Lxe–2x – 2Lxe–2x

= – 3Le–2x = e–2x

Hence

– 3L = 1 or L = –

So a PI is

yp = – xe–2x

and the GS of the equation is

y = yc + yp = Aex + Be–2x – x e–2x

c) ii) y´´– 4y´+ 4y = e–2x

yc = (Ax + B)e2x

This time we can use yp = Le–2x as a guess for the PI giving in the equation:

4Le–2x – 4(–2Le–2x) + 4Le–2x = e–2x

from which

16L = 1 or L =

and

yp = e–2x

and the GS is

y = yc + yp = (Ax + B)e2x + e–2x

c) iii) y´´+ 4y´+ 5y = e–2x

yc = e–2x (Acos x + B sin x)

With the trial yp = Le–2x (Of course the occurrence of e–2x in the CF is not a problem here since it is accompanied by cos and sin terms) a PI is found to be

yp = e–2x

So the GS is

y = yc + yp = e–2x (Acos x + B sin x) + e–2x

c) iv) y´´+ 4y = x + 1

yc = Acos 2x + B sin 2x

A PI is found to be

yp = e–2x

So the GS is

y = yc + yp = Acos 2x + B sin 2x + e–2x

c) v) y´´– 9y = e–2x

The CF is

yc = Ae3x + Be–3x

A PI is

yp = – e–2x

So the GS is

y = yc + yp = Ae3x + Be–3x – e–2x

c) vi) y´´+ y = e–2x

yc = Acos x + Bsin x

A PI is

yp = e–2x

So the GS is

y = yc + yp = Acos x + Bsin x + e–2x

d) i) y´´+ y´– 2y = 2 sin x

yc = Aex + Be–2x

For a particular integral the trial yp = L cos x + M sin x is suggested by the RHS:

yp'= – L sin x + M cos x

yp'' = – L cos x – Msin x

so

yp'' + yp' –2 yp = –L cos x – M sin x – L sin x + M cos x – 2L cos x – 2 M sinx

= (– 3L + M) cos x + (– L – 3 M) sin x = 2 sin x

Therefore

– 3L + M = 0 or M = 3L

and

– L – 3M = 2

from which we get

L = – and M = –

giving a PI

yp = – cos x – sin x

and the GS of the equation is

y = yc + yp = Aex + Be–2x – cos x – sin x

d) ii) y´´– 4y´+ 4y = 2 sin x

yc = (Ax + B)e2x

yp = Lcos x + M sin x as a guess for the PI gives this time:

(3L – 4M)cos x + (4L + 3M)sin x = 2 sin x

from which

3L – 4M = 0

4L + 3M = 2

giving

L = M =

so

yp = cos x + sin x

and the GS is

y = yc + yp = (Ax + B)e2x + cos x + sin x

d) iii) y´´+ 4y´+ 5y = 2 sin x

yc = e–2x (Acos x + B sin x)

With the trial yp = Lcos x + M sin x a PI is found to be

yp = – cos x + sin x

So the GS is

y = yc + yp = e–2x (Acos x + B sin x) – cos x + sin x

d) iv) y´´+ 4y = 2 sin x

yc = Acos 2x + B sin 2x

A PI is found to be

yp = sin x

So the GS is

y = yc + yp = Acos 2x + B sin 2x + sin x

NB In this case, as there is no y' term in the DE we could just take yp =Lsin x as a trial.

d) v) y´´– 9y = 2 sin x

The CF is

yc = Ae3x + Be–3x

A PI is

yp = – sin x

So the GS is

y = yc + yp = Ae3x + Be–3x – sin x

d) vi) y´´+ y = 2 sin x

yc = Acos x + Bsin x

In this case the RHS is included in the CF and so we have to take a trial of the form

y = x(Lcos x+ M sin x)

Substituting in the equation produces on simplification

2M cos x – 2Lsin x = 2 sin x

from which

L = – 1, M = 0

so a PI is

yp = – x cos x

So the GS is

y = yc + yp = Acos x + Bsin x – x cos x

8. Solve the initial value problem y(0) = 0, y´(0) = 0 for each of the equations solved in Q7 a). If you feel really keen press on with the other questions - you can check your answers by substituting into the equation.

Solution

a) i) y´´+ y´– 2y = 2 y(0) = y'(0) =0

The GS of the equation is

y = Aex + Be–2x – 1

y(0) = 0 gives A + B – 1 = 0

y'(0) = 0 gives A – 2B = 0

Solving these give

A = B =

so the required solution is

y = ex + e–2x – 1

a) ii) y´´– 4y´+ 4y = 2 y(0) = y'(0) = 0

The GS is

y = (Ax + B)e2x +

y(0) = 0 gives B + = 0 or B = –

y'(0) = 0 gives A + 2B = 0 or A = – 2B = 1

So the solution is

y = e2x +

a) iii) y´´+ 4y´+ 5y = 2 y(0) = y'(0) = 0

The GS is

y = e–2x (Acos x + B sin x) +

The ICs give in this case

A = – and B = 2A = –

So the solution is

y = – e–2x +

a) iv) y´´+ 4y = 2 y(0) = y'(0) = 0

The GS is

y = Acos 2x + B sin 2x +

The ICs give

A = – and B = 0

So the solution is

y =

a) v) y´´– 9y = 2 y(0) = y'(0) = 0

The GS is

y = Ae3x + Be–3x –

The ICs give, respectively

A + B =

A – B = 0

from which

A = B =

giving a particular solution of

y =

a) vi) y´´+ y = 2 y(0) = y'(0) = 0

The GS is

y = Acos x + Bsin x + 2

The ICs give A = – 2 and B = 0, so the solution is

y = 2(1 – cos x)

9. Solve the boundary value problem y(0) = y(1) = 0 for each of the equations solved in Q7 a). Again press on with the rest of Q7 if you need more practice.

Solution

Boundary value problems are always that bit more difficult to solve, because they usually lead to more complicated equations – but in principle there is not a lot of difference.

a) i) y´´+ y´– 2y = 2 y(0) = y(1) = 0

The GS of the equation is

y = Aex + Be–2x – 1

y(0) = 0 gives A + B – 1 = 0

y(1) = 0 gives eA + 2e–2B – 1 = 0

or

A + B = 1

A + e–3 B = e–1

Subtracting to remove A gives

B = = =

Then

A = 1 – B =

- all excellent practice in algebra!! The required solution is finally

y = ex + e–2x – 1

a) ii) y´´– 4y´+ 4y = 2 y(0) = y(1) = 0

The GS is

y = (Ax + B)e2x +

y(0) = 0 gives B = –

y(1) = 0 gives (A + B)e2 + = 0 from which

A =

So the solution is

y = e2x +

a) iii) y´´+ 4y´+ 5y = 2 y(0) = y(1) = 0

The GS is

y = e–2x (Acos x + B sin x) +

The BCs give

A + = 0

e–2 (Acos 1 + B sin 1) + = 0

from which we find

A = – and B =

and so

y = e–2x +

a) iv) y´´+ 4y = 2 y(0) = y(1) = 0

The GS is

y = Acos 2x + B sin 2x +

The BCs give

A = – and B =

So the solution is

y = – cos 2x + sin 2x +

a) v) y´´– 9y = 2 y(0) = y(1) = 0

The GS is

y = Ae3x + Be–3x –

The BCs give, respectively

A + B =

Ae3 + Be–3 =

or

A + Be–6 =

A tour de force in algebra should then give you

B = =

A = =

giving a solution of

y = e3x + e–3x –

a) vi) y´´+ y = 2 y(0) = y(1) = 0

The GS is

y = Acos x + Bsin x + 2

The BCs give A = – 2 and B = 2

So

y = – 2 cos x + 2 sin x + 2

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