Introduction to Differential Equations

1. Second order; linear

2. Third order; nonlinear because of (dy/dx)4

3. Fourth order; linear

4. Second order; nonlinear because of cos(r + u)

5. Second order; nonlinear because of (dy/dx)2 or
1+(dy/dx)2

6. Second order; nonlinear because of R2

7. Third order; linear

8. Second order; nonlinear because of ˙x2

9. Writing the differential equation in the form x(dy/dx) + y2 = 1, we see that it is nonlinear in y because of y2.

However, writing it in the form (y2 − 1)(dx/dy) + x = 0, we see that it is linear in x.

10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linear in v. However,

writing it in the form (v + uv − ueu)(du/dv) + u = 0, we see that it is nonlinear in u.

11. From y = e−x/2 we obtain y
= −12 e−x/2. Then 2y
+ y = −e−x/2 + e−x/2 = 0.

12. From y = 65 − 65 e−20t we obtain dy/dt = 24e−20t

, so that

dy

dt + 20y = 24e−20t + 20 65 − 65e−20t

= 24.

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

y

− 6y
+ 13y = 0.

14. From y = − cos x ln(sec x + tan x) we obtain y
= −1 + sin x ln(sec x + tan x) and

y

= tan x + cos x ln(sec x + tan x). Then y

+ y = tan x.

15. The domain of the function, found by solving x + 2 ≥ 0, is [−2,∞). From y
= 1 + 2(x + 2)−1/2 we have

(y − x)y
= (y − x)[1 + (2(x + 2)−1/2] = y − x + 2(y − x)(x + 2)−1/2 = y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2 = y − x + 8(x + 2)1/2(x + 2)−1/2 = y − x + 8.