Evaluación de la Reconstrucción con Secuencias Simuladas

u(t) f (t) dt + C, or

x(t) = 1 u(t)



u(t) f (t) dt + C

u(t), (4.13)

which is the general solution to (4.1).

Thus, the key to the method is finding an integrating factor, a function u that satisfies equation (4.12). If we expand both sides of (4.12), this becomes

ux^+ u^x = ux^− aux.

Subtracting ux^from each side, this becomes u^x = −aux, which will be satisfied if

u^= −au. (4.14)

This is a linear homogeneous equation, and, as we saw earlier in (4.4), a particular solution is given by

u(t) = e^{−a(t) dt}. (4.15)

(Notice that we do not need the constant A that appears in (4.4) because we only need one particular solution. Any solution to (4.14) will do for the present purpose.) Summary of the method

We have found a general method of solving arbitrary linear equations.

The equation

x^= ax + f. (4.16)

can be solved using the following four steps.

1. Rewrite the equation as

x^− ax = f.

2. Multiply by the integrating factor

u(t) = e^{−a(t) dt}, so that the equation becomes

(ux)^= u(x^− ax) = u f. (4.17) 3. Integrate this equation to obtain

u(t)x(t) =



u(t) f (t) dt + C.

4. Solve for x(t).

After you have found the integrating factor u in step 2, it is always a good idea to check that equation (4.17) is satisfied.

Let’s look at some examples.

E x a m p l e 4 . 1 8 Find the general solution to the equation x^= x + e^−t.

Let’s go about this very carefully. The first thing to do is to bring the term involving x to the left-hand side,

x^− x = e^−t. (4.19)

Next, since a(t) = 1, the integrating factor is u(t) = e^{−1 dt} = e^−t. Multiply equation (4.19) by the integrating factor, getting

e^−t(x^− x) = e^−2t. (4.20)

Verify that the left-hand side of (4.20) is the derivative of the product u(t)x(t) = e^−tx(t), or

[e^−tx(t)]^= e^−t(x^− x) = e^−2t. (4.21) We can now integrate both sides of (4.21),

e^−tx(t) =



e^−2tdt

= −1

2e^−2t+ C.

Finally, we solve for x by multiplying both sides by e^t, getting x(t) = −1

2e^−t+ Ce^t. (4.22)

●

E x a m p l e 4 . 2 3 Find the general solution of

x^= x sin t + 2te^{− cos t} and the particular solution that satisfies x(0) = 1.

This equation is more clearly in the linear form of (4.16) if we rewrite it as x^= (sin t)x+2te^{− cos t}. Again we start to find the solution by rewriting the equation as

x^− x sin t = 2te^{− cos t}. This time a(t) = sin t, so the integrating factor is

u(t) = e^{−sin t dt} = e^{cos t}. Multiplying by u, we get

[e^{cos t}x(t)]^= e^{cos t}(x^− x sin t) = 2t.

2.4 Linear Equations 51

After integrating both sides we have

x(t)e^{cos t} = 2



t dt= t²+ C.

Therefore, the general solution is

x(t) = (t²+ C)e^{− cos t}. (4.24) The particular solution we want satisfies x(0) = 1, so

1= Ce⁻¹ or C = e.

Thus the solution to the initial value problem is

x(t) = (t²+ e)e^{− cos t}. ●

E x a m p l e 4 . 2 5 Find the general solution to

x^= x tan t + sin t, and find the particular solution that satisfies x(0) = 2.

Rewrite the equation as

x^− x tan t = sin t.

Then a(t) = tan t, so an integrating factor is

u(t) = e^{−tan t dt} = e^{ln(cos t)}= cos t.

Multiplying by the integrating factor, we get [x cos t]^= cos t

x^− x tan t

= cos t sin t, so

x(t) cos t =



cos t sin t dt = −cos²t 2 + C.

Finally, we divide by cos t to get

x(t) = −cos t

2 + C

cos t. (4.26)

This is the general solution. To find the particular solution with x(0) = 2, we substitute this into the formula for the general solution and compute that C = 5/2.

Thus our particular solution is

x(t) = −cos t

2 + 5

2 cos t. ●

An alternate solution method

There is another method of solving linear equations that you might find easier to remember and use. Let’s begin with an example.

E x a m p l e 4 . 2 7 Find the general solution of

y^= −2y + 3. (4.28)

First, the solution to the associated homogeneous equation, y_h^ = −2yhis yh = Ce^−2t. Replace the constant in the homogeneous solution with v = v(t), a yet to be determined function of t, so

y(t) = v(t)e^−2t. (4.29)

Then we substitute this expression for y into the inhomogeneous equation (4.28) and solve forv.

(ve^−2t)^= −2(ve^−2t) + 3

−2ve^−2t + v^e^−2t = −2ve^−2t+ 3 v^= 3e^2t

v = 3 2e^2t+ C

(4.30)

Finally, substitute this last result into equation (4.29) to obtain the general solution of equation (4.28).

y=

3 2e^2t+ C



e^−2t = 3

2+ Ce^−2t ●

Notice that the derivation in (4.30) left us with a formula forv^, which we only needed to integrate to findv. It is fair to ask if this always happens. Let’s look at the general case.

We want to solve the linear equation

y^= a(t)y + f (t). (4.31)

We start by finding a particular solution to the associated homogeneous equation

y_h^ = a(t)yh. (4.32)

According to (4.4), a solution is

y_h(t) = e^{a(t) dt}. (4.33)

Notice that because of its exponential form the function y_h(t) is never equal to zero.

Hence we can safely divide by it. If y(t) is any solution to (4.31), we can define v(t) = y(t)

yh(t), so that y(t) = v(t)yh(t).

This is the key idea. We write an arbitrary solution to (4.31) in the form y(t) = v(t)yh(t). The function v is as yet unknown. It is what is sometimes called a variable parameter, and this method is called variation of parameters. To solve for v we substitute y= v(t)yh(t) into the differential equation (4.31). We get

(vyh)^= a(vyh) + f or vyh^ + v^y_h = avyh + f.

2.4 Linear Equations 53 Remember that yh is a solution of the homogeneous equation (4.32). Hence, y_h^ = ay_h, and proceeding, we get

avyh + v^y_h = avyh+ f, v^yh = f,

v^= f yh.

(4.34)

From this, we can compute v by integration. The solution is the product y = v(t)yh(t).

Here is a list of the steps in the method of variation of parameters for solving

y^= ay + f. (4.35)

1. A particular solution to the associated homogeneous equation y_h^ = ayh is

y_h(t) = e^{a(t) dt}

2. Substitute y= vyhinto equation (4.35) to findv, or remember that v^= f

y_h (4.36)

3. Write down the general solution y(t) = v(t)yh(t).

In step 2 you have a choice. Many people find it easier to substitute y = vyh into the equation than to remember the exact formula forv.

E x a m p l e 4 . 3 7 Use variation of parameters to find the general solution of

x^= x tan t + sin t, (4.38)

which we solved in Example 4.25.

The associated homogeneous equation is x_h^ = xhtan t, which has solution x_h(t) = 1/ cos t.

Next we look for a solution of the form x = vxh = v/ cos t. Substituting x into equation (4.38) and using the homogeneous equation, we get

v cos t

_

= v

cos t sin t cos t + sin t v^cos t+ v sin t

cos²t = v

cos t sin t cos t + sin t v^

cos t = sin t v^ = sin t cos t.

The last equation can be integrated, finding that v(t) =



sin t cos t dt = −cos²t 2 + C.

Finally, our solution is x(t) = v(t)xh(t) =



−cos²t

2 + C 

cos t = −cos t

2 + C

cos t,

which agrees with our previous answer. ●

Structure of the solution

The method of variation of parameters shows that an arbitrary solution to the inho-mogeneous linear equation

y^= ay + f has the form

y(t) = v(t)yh(t), where

y_h(t) = e^{a(t) dt}

is a particular solution to the associated homogeneous equation and where, accord-ing to (4.34),

v^(t) = f (t)/yh(t) = f (t)e^{−a(t) dt}. Performing the integration, we see that

v(t) =



f(t)e^{−a(t) dt}dt + C.

We have added the constant C to this formula to emphasize the presence of a con-stant of integration.

Hence, we can write an arbitrary solution as y(t) = v(t)yh(t)

= yh(t)



f(t)e^{−a(t) dt}dt+ Cyh(t). (4.39) Notice how the constant of integration C appears in this formula. It is the coefficient of the solution y_h to the associated inhomogeneous equation.

If we pick a particular solution y_p(t), it will be associated with a particular value of the constant C, say C_p, so that

y_p(t) = yh(t)



f(t)e^{−a(t) dt}dt+ Cpy_h(t). (4.40) Comparing (4.39) and (4.40), we see that the difference of the two solutions to the inhomogeneous equation is

y(t) − yp(t) = (C − Cp)yh(t).

Thus, the difference y− yp is a constant multiple of y_h and is itself a solution to the homogeneous equation. Furthermore, if we set A = C − Cp, we see that an arbitrary solution y can be written as

y(t) = yp(t) + Ayh(t).

Thus, we have demonstrated the following result, showing how the constant of in-tegration appears in the general solution to a linear equation.

THEOREM 4.41 Suppose that y_pis a particular solution to the inhomogeneous equation y^= a(t)y + f (t),

and that y_h is a particular solution to the associated homogeneous equation. Then every solution to the inhomogeneous equation is of the form

y(t) = yp(t) + Ayh(t), (4.42)

where A is an arbitrary constant. ■

2.4 Linear Equations 55

EXERCISES

In Exercises 1–12, find the general solution of the first-order, linear equation.

Use the integrating factor technique of this section to find the general solution.

(b) The equation y^+ y cos x = cos x is also separable.

Use the separation of variables technique to solve the equation and discuss any discrepancies (if any) be-tween this solution and the solution found in part (a).

In Exercises 14–17, find the solution of the initial value prob-lem.

14. y^= y + 2xe^2x, y(0) = 3

15. (x²+ 1)y^+ 3xy = 6x, y(0) = −1 16. (1 + t²)y^+ 4ty = (1 + t²)⁻², y(1) = 0 17. x^+ x cos t =¹₂sin 2t, x(0) = 1

In Exercises 18–21, find the solution of the initial value prob-lem. Discuss the interval of existence and provide a sketch of your solution.

18. x y^+ 2y = sin x, y(π/2) = 0

19. (2x + 3)y^= y + (2x + 3)^1/2, y(−1) = 0 20. y^= cos x − y sec x, y(0) = 1

21. (1 + t)x^+ x = cos t, x(−π/2) = 0

22. The presence of nonlinear terms prevents us from using

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