Problems 6.3 Solutions Solution C 0 - University of Utah

[Pages:4]Problems 6.3 Solutions

1. Solve the initial value problem xy + y = x, y(2) = 5.

Solution. First solve the homogeneous equation xy + y = 0, for which the variables separate: dy/y = -dx/x. This integrates to ln y = - ln x + C = ln(1/x) + C, which in turn exponentiates to y = K/x. So, we try y = u/x, y = u /x + ? ? ? in the original equation, remembering that the ? ? ? will cancel out the rest of the equation. We get

xu /x = x or u = x , which has the solution u = x2/2 + C. Thus

u xC y= = + .

x2x The initial condition gives 5 = 1 + C/2, so C = 8, and the answer is

x8 y= + .

2x

2. Solve the initial value problem: y = x(5 - y), y(0) = 1.

Solution. The variables separate to give:

dy = xdx .

5-y

Integrating both sides, we have

x2 - ln |5 - y| = + C ,

2

and exponentiating gives |5 - y| = Ke-x2/2 The initial condition leads to the equation |5 - 1| = K,

so K = 4. The ambiguity in the absolute value disappears since 5-1 is positive, so the solution is y = 5 - 4e-x2/2.

We remark that the ambiguity of the absolute value might resolve the other way. For example, if the initial condition were y(0) = 9, then we'd still have K = 4, but since 5-9 is negative, we have to evaluate |5 - y| as y-5, and the solution we get is y = 5 + 4e-x2/2. Notice that the constant function y = 5 solves the general differential equation, so it is not surprising that other solutions converge to this constant solution.

3. Solve the initial value problem (x + 1)y = 2y, y(1) = 1.

Solution. Separating variables, this becomes

dy 2dx

=

.

y x+1

Integrating both sides,

ln y = 2 ln(x + 1) + C ,

1

which exponentiates to y = K(x + 1)2, where K = eC . The initial values give 1 = K(1 + 1)2, so K = 1/4, and the solution is y = (x + 1)2/4.

4. Solve the initial value problem xy - y = x3, y(1) = 2.

Solution. First solve the homogeneous equation xy - y = 0, for which the variables separate: dy/y = dx/x. This has the solution y = Kx. So, we try y = ux in the original equation, getting

x2u = x3 or u = x ,

which has the solution u = x2/2 + C. Thus

x3 y = ux = + Cx .

2

The initial condition gives 2 = 1/2 + C, so C = 3/2, and the answer is

x2 + 3x

y=

.

2

5. Solve the initial value problem y - 2xy = ex2 , y(0) = 4.

Solution. First solve the homogeneous equation y - 2xy = 0, for which the variables separate: dy/y = 2xdx. This has the solution y = Kex2/2. So try y = uex2/2 in the original equation, getting

ex2/2u = ex2/2 , or u = 1 .

Thus u = x + C, and

y = uex2/2 = (x + C)ex2/2 .

From the initial conditions we get C = 4, so y = (x + 4)ex2/2 .

6. Solve the initial value problem: 4y + 3y = ex ,

y(0) = 7 .

Solution. First solve the homogeneous equation, which can be written as dy/y = -(3/4)dx, which has the solution y = Ke-(3/4)x. We try y = ue-(3/4)x in the original equation. The left hand side is

4y + 3y = 4(ue-(3/4)x) + 3ue-(3/4)x = 4u e-(3/4)x - 3ue-(3/4)x + 3ue-(3/4)x = 4u e-(3/4)x ,

so the original equation, in terms of u is

4u e-(3/4)x = ex

or

u = 1 e(7/4)x ,

4

which has the solution

u = 1 e(7/4)x + C 7

so that

y = ( 1 e(7/4)x + C)e-(3/4)x = 1 ex + Ce-(3/4)x .

7

7

2

The initial condition gives C = 48/7, and the solution is y = (1/7)(ex + 48e-(3/4)x).

7. Solve the initial value problem: xy - 3y = x2 ,

y(1) = 4 .

Solution. First solve the homogeneous equation: xy - 3y = 0:

dy dx =3

yx

so that

ln y = 3 ln x + C ,

which gives us y = Kx3. Now try y = ux3 and solve for u. The left hand side of the original

equation is

xy - 3y = (ux3) - 3ux3 = x(u x3 + 3ux2) - 3ux3 = x4u .

So we have to solve x4u = x2, or u = x-2, so

1 u=- +C

x

so that

y

=

1 (-

+

C )x3

=

C x3

-

x2

.

x

The initial values give 4 = C - 1, so C = 5, and our solution is

y = 5x3 - x2 .

8. Solve the initial value problem y - 2xy = ex2 , y(0) = 4.

Solution. First solve the homogeneous equation y - 2xy = 0, for which the variables separate: dy/y = 2xdx. This has the solution y = Kex2/2. So try y = uex2/2 in the original equation, getting

ex2/2u = ex2/2 , or u = 1 .

Thus u = x + C, and

y = uex2/2 = (x + C)ex2/2 .

From the initial conditions we get C = 4, so y = (x + 4)ex2/2 .

9. Solve the initial value problem: y + y = ex, y(0) = 5.

Solution. First solve the homogeneous equation y + y = 0. This has the solution y = Ke-x.We try y = ue-x in the given equation, leading to

u e-x = ex or u = e2x

which has the solution u = e2x/2 + C. Thus the general solution of our equation is

y

=

e2x (

+

C )e-x

=

ex

+ Ce-x

.

2

2

The initial conditions are y = 5 when x = 0. Put that in the above equation and solve for C to

get C = 9/2. Thus the answer is

ex + 9e-x

y=

.

2

3

10. Solve the initial value problem :

y y + = x, y(1) = 2 .

x

Solution. First, the homogeneous equation is separable:

dy y

dy dx

+ becomes = - ,

dx x

y

x

which has the solution y = K/x. For the inhomogeneous equation try y = u/x:

yu u u

y

+ x

=

x

- x2

+ x2

=x,

or u = x2. Integrate to get u = x3/3 + C and the general solution of our differential equation is

x2 C y= + .

3x Substituting the initial conditions, we arrive at C = 5/3, so the solution is

x2 5 y= + .

3 3x

4

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