1 Find the exact solution of the initial value problem ...

[Pages:7]Midterm 1

33B-1

2015 October 22

1 Find the exact solution of the initial value problem. Indicate the interval of existence.

x

y=

, y(-1) = 0.

1 + 2y

Solution. We observe that the equation is separable, and separate the variables:

(1 + 2y)y = x.

Integrate both sides in x:

(1 + 2y(x))y (x) dx = x dx +C.

The left-hand side can be handled by substituting z = y(x), dz = y (x)dx:

(1 + 2z)dz = z + z2 = y(x) + y2(x).

The right-hand side is a standard integral, so in the end we have the implicit solution

y(x) + y2(x) = x2 +C. 2

Since we must find the explicit solution, we rewrite this as

y2(x) + y(x) -

x2 +C

= 0.

2

This is a quadratic polynomial in y(x), so we solve using the quadratic formula:

-1 ?

1+4

x2 2

+C

y(x) =

.

2

We now use the initial condition y(-1) = 0 to find

0 = -1 ? 3 + 4C ,

2

2

or equivalently

1 = ? 3 + 4C.

This

forces

the

sign

to

be

positive,

and

we

obtain

C

=

-

1 2

.

Therefore

the

solution

is

11 y(x) = - +

1+4

x2 1 -

11 =- +

2x2 - 1 :

22

22

22

that is,

11 y(x) = - +

2x2 - 1.

22

The interval of existence for this solution is the largest interval containing the initial condition

x = -1 for which the solution is defined. Since 2x2 - 1 is not defined for - 1 < x < 1 , it

2

2

follows

that

the

interval

of

existence

is

(-,

-

1 2

].

1

2 Find the solution of the initial value problem. (1 + t2)y + 4ty = (1 + t2)-2, y(1) = 0.

Since this is a linear equation, we can solve this using either integrating factors or variation of parameters. Solution 1. We divide the equation through by (1 + t2) to put it in the form

y

+

4t 1 + t2 y

=

(1 + t2)-3.

Then the integrating factor is (making the substitution u = 1 + t2, du = 2tdt)

e

4t 1+t2

dt

= e2

du u

= e2 log(1+t2)

=

(1 + t2)2.

Multiplying through by the integrating factor, we have ((1 + t2)2y) = (1 + t2)-1.

Integrating in t, we find that

(1 + t2)2y = arctant +C.

Therefore

arctant +C y(t) = (1 + t2)2 .

Using the initial condition y(1) = 0, we have

0

=

arctan 1 +C

=

4

+C .

4

4

Therefore

C

=

-

4

and

the

solution

to

the

IVP

is

y(t)

=

arctan

t

-

4

(1 + t2)2

.

Solution 2. We now show how to use the method of variation of parameters. We first solve the

homogeneous equation

(1 + t2)yh + 4tyh = 0.

This we can do by separating variables:

yh yh

=

4t -1 + t2 .

Integrating and solving for yh, we find that

yh(t) = e-

4t 1+t2

dt

= (1 + t2)-2.

2

Now suppose that the solution to the inhomogeneous equation

y

+

4ty 1 + t2

=

(1 + t2)-3

is of the form

y = vyh.

Then the method of variation of parameters tells us that

f (1 + t2)-3

1

v = yh = (1 + t2)-2 = 1 + t2 .

where f is the inhomogeneous term f = (1 + t2)-3. Therefore

1 v(t) = 1 + t2 dt = arctant +C,

and so the general solution is given by

arctant +C y(t) = v(t)yh(t) = (1 + t2)2 .

Similar

calculations

as

in

solution

1

now

show

that C

=

-

4

and

therefore

y(t)

=

arctan

t

-

4

(1 + t2)2

.

3

3

Use

the

integrating

factor

?(x, y)

=

1 xy

to make the following equation into an exact equation.

Then find the general solution.

(x2y2 - 1)ydx + (1 + x2y2)xdy = 0.

Solution. Multiplying through by the integration factor gives us the equation

(xy2 - 1 )dx + ( 1 + x2y)dy = 0.

x

y

Setting

P(x, y)

=

xy2

-

1 x

and

Q(x, y)

=

1 y

+ x2y,

we

can

verify

that

P = 2xy = Q,

y

x

so this new equation is exact. Using the relations

F = P, F = Q,

x

y

we solve for F. There are two equivalent ways to do this:

(a) We integrate P in x to find

x2y2 F(x, y) = P(x, y) dx + (y) = 2 - log x + (y).

Differentiating in y, we have Since F/ y = Q, we see that

F y

=

x2y +

(y).

x2y +

(y)

=

x2y

+

1 ,

y

or

(y)

=

1. y

Therefore (y) = log y, and the general solution is given implicitly by

x2y2 F(x, y) = - log x + log y = C,

2 where C is an arbitrary constant.

(b) We integrate Q in y to find

x2y2 F(x, y) = Q(x, y) dy + (x) = + log y + (x).

2

4

Differentiating in x, we have Since F/ x = P, we see that or Therefore

F x

=

xy2 +

(x).

xy2

+

(x)

=

xy2

-

1 ,

x

1 (x) = - x .

(x) = - log x,

and the general solution is given by

x2y2 F(x, y) = + log y - log x = C,

2

where C is an arbitrary constant.

5

4 Suppose that x is a solution to the initial value problem x = x - t2 + 2t, x(0) = 1.

Show that x(t) > t2 for all t for which x is defined. There is an easy way to solve this problem, and a harder way. Solution 1 (easy). First, we show that the above equation satisfies the conditions of the uniqueness theorem for initial value problems for first-order ODEs. We set f (t, x) = x - t2 + 2t. Then f is continuous everywhere in the tx-plane, and

f =1 x is also continuous in the tx-plane, so the conditions of the uniqueness theorem are satisfied. Therefore solutions to the IVP for x = x -t2 + 2t are unique. In particular, two distinct solutions to the ODE x = x - t2 + 2t may never cross in the tx-plane. Now we observe that x1(t) = t2 solves the IVP x1 = x1 - t2 + 2t, x1(0) = 0. For x1(0) = 02 = 0, and x1(t) = 2t = t2 - t2 + 2t = x1(t) - t2 + 2t. So x1 and x are both solutions to y = f (t, y). Since x1(0) = 0 < 1 = x(0), the two solutions are distinct. By uniqueness, the solutions curves may never cross in the tx-plane; it follows then that x1(t) < x(t) for all t. That is to say, t2 < x(t) for all t, as claimed. Solution 2 (harder). Since the equation is linear one can solve it using any of the valid methods for first-order linear ODEs. With some work one can show, after some careful calculations involving multiple integrations by parts, that the solution to the IVP is given by x(t) = t2 + et. Since et > 0 for all t, it follows that x(t) > t2 for all t. (While the description of this solution may seem simpler, it is harder in the sense that it is far easier to make a mistake while trying to solve the equation explicitly.)

6

5 Find the general solution for the following differential equation. 4y + 4y + y = 0.

Solution. This is a second-order constant-coefficient linear equation. The characteristic equation

for this equation is

4 2 + 4 + 1 = 0.

Using the quadratic formula, we see that

-4 ? 4 - 4 1

=

2

=- . 2

Since the characteristic equation has repeated roots, a fundamental system for the differential

equation is given by

y1 (t )

=

e-

1 2

t

,

y2 (t )

=

t

e-

1 2

t

.

Therefore the general solution is given by a linear combination of these fundamental solutions,

that is,

y(t)

=

Ae-

1 2

t

+

Bt e-

1 2

t

,

where A and B are arbitrary real constants.

7

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