Math 432 HW 1.2 Solutions
Assigned: 1, 2, 3, 5, 8, 9, 12, 13, 15, 18, 19, 21, 22, 23, 27, 28, 30, and 31
NOTE: For #18, the text from "Note that" to the end is just for your edification.
Selected for Grading: 3, 9, 12, 23, 28
Solutions:
1. (a) We have the differential equation x(dy/dx) = 2y, the proposed solution φ(x) = x2, and the interval
I = (–∞, ∞).
φ'(x) = 2x
So xφ'(x) = x(2x) = 2x2 = 2φ(x).
So φ(x) is a solution, and it is defined on the given interval.
(b) Here we have the differential equation dy/dx + y2 = e2x + (1 – 2x)ex + x2 – 1, the proposed solution
φ(x) = ex – x, and the interval (–∞, ∞).
φ'(x) = ex – 1.
So
φ'(x) + φ(x)2 = ex – 1 + (ex – x)2
= ex – 1 + e2x – 2xex + x2
= ex(1 – 2x) + e2x – 1 + x2
= e2x + (1 – 2x)ex + x2 – 1, as needed.
That much verifies that φ(x) is a solution. And since this function is defined on all of (–∞, ∞) then we're
finished with this part.
(c) Finally, we have x2 d 2y/dx2 = 2y and φ(x) = x2 – x –1 on the interval (0, ∞).
φ'(x) = 2x + x –2
φ''(x) = 2 – 2x –3
So x2φ''(x) = x2(2 – 2x –3) = 2x2 – 2x –1 = 2(x2 – x –1) = 2φ(x).
Thus this function φ(x) is indeed a solution to the given differential equation.
And, since the function is defined on the interval (0, ∞), then we've finished with this part too.
NOTE: This function is also a solution to the differential equation on the interval (–∞, 0).
2. (a) The differential equation is dy/dx = –1/(2y) and the solution is supposedly given implicitly by the
relation y2 + x – 3 = 0 on the interval (–∞, 3).
Differentiating both sides of the relation gives 2y(dy/dx) + 1 = 0.
Solving this for dy/dx gives dy/dx = –1/(2y), so this relation does give a solution implicitly.
What about the interval? Take an algebraic look at the relation y2 + x – 3 = 0. Solving this for x we
get x = 3 – y2. Since y2 ≥ 0, then this shows that x ≤ 3. Since our interval must be an open one, we
throw out the endpoint 3 and are left with the interval x < 3 or (–∞, 3).
The following graph of the given relation make things a little easier to see.
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y
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x
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(b) The relation xy3 – xy3sin x = 1 can be solved for y.
xy3(1 – sin x) = 1
y3 = 1/[x(1 – sin x)] {Note that making that step assumes that x ≠ 0 and that sin x ≠ 1.}
y = {1/[x(1 – sin x)]}1/3 = [x(1 – sin x)] –1/3
Differentiating:
This much verifies that the implicitly given y is a solution to the differential equation. As for the
interval of definition, any interval that does not include x = 0 and that does not include any x-value for
which sin x = 0 would do. The given interval (0, π/2) is one such interval.
3. Given: y = sin x + x2.
dy/dx = cos x + 2x and 2y/dx2 = – sin x + 2.
So d 2y/dx2 + y = – sin x + 2 + sin x + x2 = 2 + x2 = x2 + 2.
Yes, this is a solution to the given differential equation. [On the interval (–∞, ∞).]
5. Given: θ = 2e 3t – e 2t.
dθ/dt = 6e 3t – 2e 2t and d 2θ/dt 2 = 18e 3t – 4e 2t.
So
d 2θ/dt 2 – θ(dθ/dt) + 3θ = 18e 3t – 4e 2t – (2e 3t – e 2t)(6e 3t – 2e 2t) + 3(2e 3t – e 2t)
= 18e 3t – 4e 2t – 12e 6t + 10e 5t – 2e 4t + 6e 3t – 3e 2t
= –12e6t + 10e 5t – 2e 4t + 24 e 3t – 7e 2t
≠ e 2t
No, this is not a solution to the given differential equation.
8. Given: y = 3 sin 2x + e –x.
y' = 6 cos 2x – e–x and y'' = –12 sin 2x + e–x.
So y'' + 4y = –12 sin 2x + e–x + 4(3 sin 2x + e –x) = 5e–x.
Yes, this is a solution to the given differential equation.
9. Given: x2 + y2 = 4.
Differentiating: 2x + 2y∙y' = 0.
Solving for y': y' = –x/y.
No, this is not a solution to the differential equation dy/dx = x/y.
12. Given: x2 – sin(x + y) = 1.
Differentiating: 2x – cos(x + y)(1 + y') = 0.
Solving for y':
1 + y' = 2x/cos(x + y)
y' = 2x/cos(x + y) – 1 = 2x sec(x + y) – 1.
Yes, this relation is an implicit solution to the given differential equation.
13. Differentiate both sides of the equation sin y + xy – x3 = 2 and you get
(cos y) y' + xy' + y – 3x2 = 0.
Solving this equation for y' gives y' = (3x2 – y)(cos y + x).
Differentiating both sides of this second equation you get
(cos y) y'' + y'(–sin y) y' + xy'' + y' + y' – 6x = 0.
Solving this for y'':
(cos y + x) y'' = 6x – 2y' + (sin y)(y')2
So the answer is "yes".
15. Given: φ(x) = 2/(1 – cex) = 2(1 – cex) –1 for some arbitrary constant, c.
φ'(x) = –2(1 – cex)–2(–cex) = 2cex/(1 – cex)2
And so
So each such φ(x) is indeed a solution to the given differential equation.
Here are the requested graphs.
y
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c=1
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c=0
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c = –2
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c = –1
x
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18. Given: c > 0; φ(x) = (c2 – x2) –1; and dy/dx – 3y = –3, y(0) = 1/c2 on (–c, c).
Differentiating gives φ'(x) = –(c2 – x2) –2(–2x) = 2x/(c2 – x2) 2 = 2xy2.
This much shows that this φ(x) is a solution to the differential equation on some interval.
It is also a solution to the initial-value problem since φ(0) = (c2 – 02) –1 = 1/c2.
Finally, note that φ(x) is defined on any interval that excludes both x = c and x = –c, for example the
given interval.
19. The equation (dy/dx)2 + y2 + 4 = 0 can't have any real-valued solution because if there were a solution then
we'd have a square, namely (dy/dx)2, equal to a negative number, namely – 4 – y2. Since that can't be, then
no such solution could exist.
21. For both parts (a) and (b) we start with φ(x) = xm for some yet-to-be-determined m. Also for both parts
we'll need the first and second derivatives of φ(x), so I'll calculate them here:
φ'(x) = mxm – 1 and φ''(x) = m(m – 1)xm – 2.
(a) I'll just make the substitution and see whether it tells me what m must be equal to.
3x2φ''(x) + 11xφ'(x) – 3φ(x) = 3x2m(m – 1)xm – 2 + 11xmxm – 1 – 3xm
= xm[3m(m – 1) + 11m – 3]
= xm(3m2 + 8m – 3)
For that to be equal to zero, we'd need for that second factor to equal zero:
3m2 + 8m – 3 = 0
(3m – 1)(m + 3) = 0
m = 1/3, –3.
So there are two possible solutions: y = x1/3 and y = x–3.
{Note that this second one's interval would have to exclude x = 0.}
(b) x2φ''(x) – xφ'(x) – 5φ(x) = x2m(m – 1)xm – 2 – xmxm – 1 – 5xm
= xm[m(m – 1) – m – 5]
= xm(m2 – 2m – 5)
So, using the quadratic formula, we get that
So there are two possible solutions:
.
and
.
{Here again, that second solution is defined on either (–∞, 0) or (0, ∞), not for all real numbers.}
22. Given: the differential equation d 2y/dx2 + dy/dx – 2y = 0 and a proposed solution φ(x) = c1ex + c2e –2x.
Verification:
φ'(x) = c1ex – 2c2e –2x
φ''(x) = c1ex + 4c2e –2x
φ''(x) + φ'(x) – 2φ(x) = c1ex + 4c2e –2x + c1ex – 2c2e –2x – 2(c1ex + c2e –2x)
= ex(c1 + c1 – 2c1) + e –2x(4c2 – 2c2 – 2c2) = 0 + 0 = 0
(a) From the initial conditions y(0) = 2 and y'(0) = 1 we get two equations:
φ(0) = c1 + c2 = 2 and
φ'(0) = c1 – 2c2 = 1
Solving the first for c1 gives c1 = 2 – c2.
Substituting that into the second equation gives 2 – c2 – 2c2 = 1, and hence c2 = 1/3.
Then c1 = 2 – 1/3 = 5/3.
Answer: c1 = 5/3, c2 = 1/3.
(b) From the initial conditions y(1) = 1 and y'(1) = 0 we get two equations:
φ(1) = c1e + c2e –2 = 1 and
φ'(1) = c1e – 2c2e –2 = 0
Solving the second for c1 gives c1 = 2c2e –3.
Substituting that into the first equation gives 2c2e –2 + c2e –2 = 1, which yields c2 = e2/3.
That means that c1 = 2(e2/3)e –3 = 2/(3e).
Answer: c1 = 2/(3e), c2 = e2/3.
23. For the initial-value problem dy/dx = y 4 – x 4, y(0) = 7 we have
f (x, y) = y 4 – x 4 and ∂f /∂y = 4y3, both of which are also continuous on the entire plane.
So Theorem 1 does imply the existence of a unique solution.
27. For the initial-value problem y(dy/dx) = x, y(1) = 0 we first solve for dy/dx: dy/dx = x/y.
But the function f (x, y) is not even defined for the initial-value point (1, 0).
So Theorem 1 does not imply the existence of a unique solution.
NOTE: Theorem 1 does not say that there is not a unique solution to the initial-value problem in this case;
it merely does not say there is one. {Difference: I say there isn't vs. I don't say there is.}
28. This time we have the initial-value problem
, y(2) = 1.
The function f (x, y) is continuous in an open rectangle containing (2, 1).
The first partial
, however, is not continuous at any point whose y-coördinate is 1.
So Theorem 1 says nothing – it does not guarantee the existence and uniqueness of a solution.
29. (a) First consider the constant function φ1(x) ≡ 0.
This function's first derivative is φ1'(x) ≡ 0 as well and so we do get φ1'(x) = 3φ1(x)2/3.
Thus this function is a solution to the differential equation.
And it certainly satisfies the initial condition: φ1(2) = 0.
So y = φ1(x) is a solution to the IVP.
For the second function φ2(x) = (x – 2)3 we have φ2'(x) = 3(x – 2)2 = 3φ2(x)2/3.
And φ2(2) = (2 – 2)3 = 0.
So this too is a solution to the given IVP.
(b) The very similar IVP y' = 3y2/3, y(0) = 10–7 does have a unique solution in a neighborhood of x = 0.
You can be sure of this by Theorem 1 since both f (x, y) = 3y2/3 and the partial ∂f /∂y = 2y–1/3 are
continuous on the rectangle R: –∞ < x < ∞, 0 < y < ∞ which contains the point (0, 10 –7).
30. First I want to check that (0, –1) is on the curve given by x + y + exy = 0.
So I'll plug in: 0 + (–1) + e0(–1) = 0 – 1 + 1 = 0. Good.
Next, for the function G(x, y) = x + y + exy we have the partial derivatives:
∂G/∂x = 1 + yexy is continuous on the entire plane, and
∂G/∂y = 1 + xexy is continuous on the entire plane as well.
Next we check the value of Gy(0, –1) = 1 + 0 = 1 ≠ 0.
So all the conditions of the Implicit Function Theorem are met.
Therefore there exists a differentiable function y = φ(x) defined on some interval I = (–δ, δ) that satisfies
the relation G(x, φ(x)) = 0 for –δ < x < δ.
31. Given: y(dy/dx) – 4x = 0.
(a) Solving for dy/dx gives dy/dx = 4x/y (which, you will note, assumes that y ≠ 0).
For any initial condition y(x0) = 0, the function f (x, y) is not continuous at (x0, 0), so Theorem 1 does
not guarantee anything.
(b) For x0 ≠ 0 and y(x0) = 0 we'd need y(x0)y'(x0) – 4x0 = 0 – 4x0 = – 4x0 to be equal to zero, which is
impossible. So the IVP couldn't possibly have a solution in any neighborhood of x0.
(c) One solution to the IVP y(dy/dx) – 4x = 0, y(0) = 0 is y = 2x.
And another is y = –2x.
Verifications:
For y = 2x, y' = 2 and so yy' – 4x = (2x)(2) – 4x = 0, and also y(0) = 0,
For y = –2x, y' = –2 and so yy' – 4x = (–2x)(–2) – 4x = 0, and also y(0) = 0.