An equation of the form dy dx (5) y = x- + f(dy/dx), where the continuously differentiable function f(t) is evaluated at t = dy/dx, is called a Clairaut equation. Interest in these equations is due to the fact that (5) has a one-parameter family of solu- tions that consist of straight lines. Further, the envelope of this family-that is, the curve whose tangent lines are given by the family is also a solution to (5) and is called the singular solution. To solve a Clairaut equation: (a) Differentiate equation (5) with respect to x and simplify to show that (6) [x+ f'(dy/dx)] = 0, d²y dx (b) From (6), conclude that dy/dx = c or f'(dy/dx) = -x. Assume that dy/dx = c and substitute back into (5) to obtain the family of straight-line solutions y = x[ y = cx + f(c). (c) Show that another solution to (5) is given parametrically by x = − f'(p), y = f(p) - pf' (p), where the parameter p dy/dx. This solution is the singular solution. (d) Use the above method to find the family of straight-line solutions and the singular solution to the equation dy dx +2 d where ƒ'(1) = f(1) dy dx 2 Here f(t) = 2t². Sketch several of the straight-line solutions along with the singular solution on the same coordinate system. Observe that the straight-line solutions are all tangent to the singular solution. (e) Repeat part (d) for the equation x(dy/dx)³-y(dy/dx)² + 2 = 0.
An equation of the form dy dx (5) y = x- + f(dy/dx), where the continuously differentiable function f(t) is evaluated at t = dy/dx, is called a Clairaut equation. Interest in these equations is due to the fact that (5) has a one-parameter family of solu- tions that consist of straight lines. Further, the envelope of this family-that is, the curve whose tangent lines are given by the family is also a solution to (5) and is called the singular solution. To solve a Clairaut equation: (a) Differentiate equation (5) with respect to x and simplify to show that (6) [x+ f'(dy/dx)] = 0, d²y dx (b) From (6), conclude that dy/dx = c or f'(dy/dx) = -x. Assume that dy/dx = c and substitute back into (5) to obtain the family of straight-line solutions y = x[ y = cx + f(c). (c) Show that another solution to (5) is given parametrically by x = − f'(p), y = f(p) - pf' (p), where the parameter p dy/dx. This solution is the singular solution. (d) Use the above method to find the family of straight-line solutions and the singular solution to the equation dy dx +2 d where ƒ'(1) = f(1) dy dx 2 Here f(t) = 2t². Sketch several of the straight-line solutions along with the singular solution on the same coordinate system. Observe that the straight-line solutions are all tangent to the singular solution. (e) Repeat part (d) for the equation x(dy/dx)³-y(dy/dx)² + 2 = 0.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![F Clairaut Equations and Singular Solutions
An equation of the form
dy
(5) y = x- + f(dy/dx),
dx
where the continuously differentiable function f(t) is evaluated at t = dy/dx, is called a Clairaut
equation. Interest in these equations is due to the fact that (5) has a one-parameter family of solu-
tions that consist of straight lines. Further, the envelope of this family that is, the curve whose
tangent lines are given by the family is also a solution to (5) and is called the singular solution.
To solve a Clairaut equation:
(a) Differentiate equation (5) with respect to x and simplify to show that
(6)
[x+ f'(dy/dx)] = 0,
d²y
dx
y = x
(b) From (6), conclude that dy/dx = cor f' (dy/dx) = -x. Assume that dy/dx = c and
substitute back into (5) to obtain the family of straight-line solutions
y = cx + f(c).
(c) Show that another solution to (5) is given parametrically by
x = -f' (p),
y = f(p) - pf' (p),
where the parameter p = dy/dx. This solution is the singular solution.
(d) Use the above method to find the family of straight-line solutions and the singular solution
to the equation
dy
dx
+21
=
dy
dx
d
dt
2
where ƒ'(0) = f(0).
f'
Here f(t)
21². Sketch several of the straight-line solutions along with the singular
solution on the same coordinate system. Observe that the straight-line solutions are all
tangent to the singular solution.
(e) Repeat part (d) for the equation
x(dy/dx)³-y(dy/dx)² + 2 = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff9bb4c1a-af37-4a70-bd33-011df9211e08%2F7bd75d16-f6ce-416b-bdb0-d4a9f691ffe4%2F74amh0g_processed.png&w=3840&q=75)
Transcribed Image Text:F Clairaut Equations and Singular Solutions
An equation of the form
dy
(5) y = x- + f(dy/dx),
dx
where the continuously differentiable function f(t) is evaluated at t = dy/dx, is called a Clairaut
equation. Interest in these equations is due to the fact that (5) has a one-parameter family of solu-
tions that consist of straight lines. Further, the envelope of this family that is, the curve whose
tangent lines are given by the family is also a solution to (5) and is called the singular solution.
To solve a Clairaut equation:
(a) Differentiate equation (5) with respect to x and simplify to show that
(6)
[x+ f'(dy/dx)] = 0,
d²y
dx
y = x
(b) From (6), conclude that dy/dx = cor f' (dy/dx) = -x. Assume that dy/dx = c and
substitute back into (5) to obtain the family of straight-line solutions
y = cx + f(c).
(c) Show that another solution to (5) is given parametrically by
x = -f' (p),
y = f(p) - pf' (p),
where the parameter p = dy/dx. This solution is the singular solution.
(d) Use the above method to find the family of straight-line solutions and the singular solution
to the equation
dy
dx
+21
=
dy
dx
d
dt
2
where ƒ'(0) = f(0).
f'
Here f(t)
21². Sketch several of the straight-line solutions along with the singular
solution on the same coordinate system. Observe that the straight-line solutions are all
tangent to the singular solution.
(e) Repeat part (d) for the equation
x(dy/dx)³-y(dy/dx)² + 2 = 0.
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