Find the order and degree of the following differential equations:(i)

1.	Find the order and degree of the following differential equations:(i) dy/dx + y = x2(ii) y’ + y2 + y3 = 0(iii) y’' + 3y'2 + y3 = 0(iv) (d2y/dx2) + x = √(y + (dy/dx)) (v) (d2y/dx2) - y + ((dy/dx) + (d3y/dx3))3/2 = 0
Answer» (i) Order = 1, Degree = 1 [Here, we have (dy/dx)'] (ii) Order = 1, Degree = 1 [We have (dy/dx)'] (iii) We have (d²y/dx²)'; So, order = 2; Degree = 1 (iv) Squaring both sides, [(d²y/dx²) + x]² = y + (dy/dx) We have (d²y/dx²)² ∴ Order = 2 Degree = 2 (v) Squaring both sides, we get ((d²y/dx²) - y)² = ((dy/dx) + (d³y/dx³))³ = 0 We have (d³y/dx³)³ Order = 3; Degree = 3

Find the order and degree of the following differential equations:(i) dy/dx + y = x2(ii) y’ + y2 + y3 = 0(iii) y’' + 3y'2 + y3 = 0(iv) (d2y/dx2) + x = √(y + (dy/dx)) (v) (d2y/dx2) - y + ((dy/dx) + (d3y/dx3))3/2 = 0

Answer»

(i) Order = 1, Degree = 1

[Here, we have (dy/dx)']

(ii) Order = 1, Degree = 1

[We have (dy/dx)']

(iii) We have (d²y/dx²)'; So, order = 2; Degree = 1

(iv) Squaring both sides, [(d²y/dx²) + x]² = y + (dy/dx)

We have (d²y/dx²)²

∴ Order = 2

Degree = 2

(v) Squaring both sides, we get

((d²y/dx²) - y)² = ((dy/dx) + (d³y/dx³))³ = 0

We have (d³y/dx³)³

Order = 3; Degree = 3