Exercises 1.1
1 Introduction to Differential Equations 16. Implicitly differentiating the solution we obtain
dy dy
y
−2x 2
− 4xy + 2y = 0 =⇒ −x2 dy − 2xy dx + y dy = 0 4
dx dx
Exercises 1.1 =⇒ 2xy dx + (x2 − y)dy = 0. 2
Using the quadratic formula to solve y 2 − 2x2 y − 1 = 0 for y, we get x
√ √ -4 -2 2 4
1. Second-order; linear. y = 2x2 ± 4x4 + 4 /2 = x2 ± x4 + 1 . Thus, two explicit solutions are
√ √ -2
2. Third-order; nonlinear because of (dy/dx)4 . y1 = x + x4 + 1 and y2 = x − x4 + 1 . Both solutions are defined on
2 2
3. The differential equation is first-order. Writing it in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in (−∞, ∞). The graph of y1 (x) is solid and the graph of y2 is dashed. -4
y because of y 2 . However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is linear in x. 17. Differentiating P = c1 et / (1 + c1 et ) we obtain
u
4. The differential equation is first-order. Writing it in the form u(dv/du) + (1 + u)v = ue we see that it is linear dP (1 + c1 et ) c1 et − c1 et · c1 et
= 2
in v. However, writing it in the form (v + uv − ueu )(du/dv) + u = 0, we see that it is nonlinear in u. dt (1 + c1 et )
5. Fourth-order; linear c1 et [(1 + c1 et ) − c1 et ]
= = P (1 − P ).
6. Second-order; nonlinear because of cos(r + u) 1 + c1 et 1 + c1 et
7. Second-order; nonlinear because of 1 + (dy/dx)2 x
18. Differentiating y = e−x et dt + c1 e−x we obtain
2 2 2
2
8. Second-order; nonlinear because of 1/R 0
x x
9. Third-order; linear y = e−x ex − 2xe−x
2 2 2
et dt − 2c1 xe−x = 1 − 2xe−x
2 2 2
et dt − 2c1 xe−x .
2 2
0 0
10. Second-order; nonlinear because of ẋ2
Substituting into the differential equation, we have
11. From y = e−x/2 we obtain y = − 12 e−x/2 . Then 2y + y = −e−x/2 + e−x/2 = 0. x x
y + 2xy = 1 − 2xe−x et dt − 2c1 xe−x + 2xe−x et dt + 2c1 xe−x = 1.
2 2 2 2 2 2
12. From y = 6
5 − 65 e−20t we obtain dy/dt = 24e−20t , so that
0 0
dy 6 6 −20t
+ 20y = 24e−20t + 20 − e = 24. dy d2 y
dt 5 5 19. From y = c1 e2x + c2 xe2x we obtain = (2c1 + c2 )e2x + 2c2 xe2x and = (4c1 + 4c2 )e2x + 4c2 xe2x , so that
dx dx2
13. From y = e3x cos 2x we obtain y = 3e3x cos 2x − 2e3x sin 2x and y = 5e3x cos 2x − 12e3x sin 2x, so that d2 y dy
y − 6y + 13y = 0. −4 + 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = 0.
dx2 dx
14. From y = − cos x ln(sec x + tan x) we obtain y = −1 + sin x ln(sec x + tan x) and
20. From y = c1 x−1 + c2 x + c3 x ln x + 4x2 we obtain
y = tan x + cos x ln(sec x + tan x). Then y + y = tan x.
dy
15. Writing ln(2X − 1) − ln(X − 1) = t and differentiating implicitly we obtain = −c1 x−2 + c2 + c3 + c3 ln x + 8x,
X dx
2 dX 1 dX d2 y
− =1 4 = 2c1 x−3 + c3 x−1 + 8,
2X − 1 dt X − 1 dt dx2
2
and
2 1 dX
− =1
2X − 1 X − 1 dt t d3 y
2X − 2 − 2X + 1 dX
-4 -2 2 4
= −6c1 x−4 − c3 x−2 ,
=1 dx3
-2
(2X − 1)(X − 1) dt so that
dX -4 d3 y d2 y dy
= −(2X − 1)(X − 1) = (X − 1)(1 − 2X). x3 + 2x2 2 − x +y
dt dx3 dx dx
= (−6c1 + 4c1 + c1 + c1 )x−1 + (−c3 + 2c3 − c2 − c3 + c2 )x
Exponentiating both sides of the implicit solution we obtain
+ (−c3 + c3 )x ln x + (16 − 8 + 4)x2
2X − 1 et − 1
= et =⇒ 2X − 1 = Xet − et =⇒ (et − 1) = (et − 2)X =⇒ X = t . = 12x2 .
X −1 e −2
Solving et − 2 = 0 we get t = ln 2. Thus, the solution is defined on (−∞, ln 2) or on (ln 2, ∞). The graph of the −x2 , x<0 −2x, x<0
21. From y = we obtain y = so that xy − 2y = 0.
solution defined on (−∞, ln 2) is dashed, and the graph of the solution defined on (ln 2, ∞) is solid. x2 , x≥0 2x, x≥0
1 2
1 Introduction to Differential Equations 16. Implicitly differentiating the solution we obtain
dy dy
y
−2x 2
− 4xy + 2y = 0 =⇒ −x2 dy − 2xy dx + y dy = 0 4
dx dx
Exercises 1.1 =⇒ 2xy dx + (x2 − y)dy = 0. 2
Using the quadratic formula to solve y 2 − 2x2 y − 1 = 0 for y, we get x
√ √ -4 -2 2 4
1. Second-order; linear. y = 2x2 ± 4x4 + 4 /2 = x2 ± x4 + 1 . Thus, two explicit solutions are
√ √ -2
2. Third-order; nonlinear because of (dy/dx)4 . y1 = x + x4 + 1 and y2 = x − x4 + 1 . Both solutions are defined on
2 2
3. The differential equation is first-order. Writing it in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in (−∞, ∞). The graph of y1 (x) is solid and the graph of y2 is dashed. -4
y because of y 2 . However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is linear in x. 17. Differentiating P = c1 et / (1 + c1 et ) we obtain
u
4. The differential equation is first-order. Writing it in the form u(dv/du) + (1 + u)v = ue we see that it is linear dP (1 + c1 et ) c1 et − c1 et · c1 et
= 2
in v. However, writing it in the form (v + uv − ueu )(du/dv) + u = 0, we see that it is nonlinear in u. dt (1 + c1 et )
5. Fourth-order; linear c1 et [(1 + c1 et ) − c1 et ]
= = P (1 − P ).
6. Second-order; nonlinear because of cos(r + u) 1 + c1 et 1 + c1 et
7. Second-order; nonlinear because of 1 + (dy/dx)2 x
18. Differentiating y = e−x et dt + c1 e−x we obtain
2 2 2
2
8. Second-order; nonlinear because of 1/R 0
x x
9. Third-order; linear y = e−x ex − 2xe−x
2 2 2
et dt − 2c1 xe−x = 1 − 2xe−x
2 2 2
et dt − 2c1 xe−x .
2 2
0 0
10. Second-order; nonlinear because of ẋ2
Substituting into the differential equation, we have
11. From y = e−x/2 we obtain y = − 12 e−x/2 . Then 2y + y = −e−x/2 + e−x/2 = 0. x x
y + 2xy = 1 − 2xe−x et dt − 2c1 xe−x + 2xe−x et dt + 2c1 xe−x = 1.
2 2 2 2 2 2
12. From y = 6
5 − 65 e−20t we obtain dy/dt = 24e−20t , so that
0 0
dy 6 6 −20t
+ 20y = 24e−20t + 20 − e = 24. dy d2 y
dt 5 5 19. From y = c1 e2x + c2 xe2x we obtain = (2c1 + c2 )e2x + 2c2 xe2x and = (4c1 + 4c2 )e2x + 4c2 xe2x , so that
dx dx2
13. From y = e3x cos 2x we obtain y = 3e3x cos 2x − 2e3x sin 2x and y = 5e3x cos 2x − 12e3x sin 2x, so that d2 y dy
y − 6y + 13y = 0. −4 + 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = 0.
dx2 dx
14. From y = − cos x ln(sec x + tan x) we obtain y = −1 + sin x ln(sec x + tan x) and
20. From y = c1 x−1 + c2 x + c3 x ln x + 4x2 we obtain
y = tan x + cos x ln(sec x + tan x). Then y + y = tan x.
dy
15. Writing ln(2X − 1) − ln(X − 1) = t and differentiating implicitly we obtain = −c1 x−2 + c2 + c3 + c3 ln x + 8x,
X dx
2 dX 1 dX d2 y
− =1 4 = 2c1 x−3 + c3 x−1 + 8,
2X − 1 dt X − 1 dt dx2
2
and
2 1 dX
− =1
2X − 1 X − 1 dt t d3 y
2X − 2 − 2X + 1 dX
-4 -2 2 4
= −6c1 x−4 − c3 x−2 ,
=1 dx3
-2
(2X − 1)(X − 1) dt so that
dX -4 d3 y d2 y dy
= −(2X − 1)(X − 1) = (X − 1)(1 − 2X). x3 + 2x2 2 − x +y
dt dx3 dx dx
= (−6c1 + 4c1 + c1 + c1 )x−1 + (−c3 + 2c3 − c2 − c3 + c2 )x
Exponentiating both sides of the implicit solution we obtain
+ (−c3 + c3 )x ln x + (16 − 8 + 4)x2
2X − 1 et − 1
= et =⇒ 2X − 1 = Xet − et =⇒ (et − 1) = (et − 2)X =⇒ X = t . = 12x2 .
X −1 e −2
Solving et − 2 = 0 we get t = ln 2. Thus, the solution is defined on (−∞, ln 2) or on (ln 2, ∞). The graph of the −x2 , x<0 −2x, x<0
21. From y = we obtain y = so that xy − 2y = 0.
solution defined on (−∞, ln 2) is dashed, and the graph of the solution defined on (ln 2, ∞) is solid. x2 , x≥0 2x, x≥0
1 2
