FINDING ORDER AND DEGREE OF DIFFERENTIAL EQUATION

Find the order and degree of the following differential equations.

(1)  (dy/dx) + y  =  x²

(2)  y' + y²  =  x

(3)  y'' + 3 (y')² + y³

(4)  d²y/dx² + x  =  √[y + (dy/dx)]

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

(6)  y''  =  (y - (y')³)^(2/3)

(7)  y' + (y'')²  =  (x + y'')²

(8)  (dy/dx)² + x = (dx/dy) + x²

Problem 1 :

(dy/dx) + y  =  x²

Solution :

By differentiating y with respect to x, we get dy/dx and its highest exponent is 1. So, order  =  1 and degree  =  1.

Order  =  1,  Degree  =  1.

Problem 2 :

y' + y²  =  x

Solution :

By differentiating y with respect to x, we get y' and its highest exponent is 1. So, order  =  1 and degree  =  1.

Problem 3 :

y'' + (3y')² + y³

Solution :

By differentiating y with respect to x two times, we get y''. So, order  =  2 and degree  =  1.

Problem 4 :

d²y/dx² + x  =  √[y + (dy/dx)]

Solution :

d²y/dx² + x  =  √[y + (dy/dx)]

Take squares on both sides.

(d²y/dx² + x)²  =  [y + (dy/dx)]

By differentiating y with respect to x two times, we get d²y/dx². So, order  =  2 and degree  =  2.

Problem 5 :

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

Solution :

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

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

Take squares on both sides

(d²y/dx² - y)²  =  (dy/dx + d³y/dx³)³

So, order  =  3 and degree  =  3.

Problem 6 :

y''  =  (y - (y')³)^(2/3)

Solution :

y''  =  (y - (y')³)^(2/3)

Take cubes on both sides.

(y'')³  =  (y - (y')³)²

So, order is 2 and degree is 3.

Problem 7 :

y' + (y'')²  =  (x + y'')²

Solution :

y' + (y'')²  =  (x + y'')²

y' + (y'')²  =  x² + (y'')² + 2xy''

x²- 2xy''-y'  =  0

So, the order is 2 and degree is 1.

Problem 8 :

(dy/dx)² + x  =  (dx/dy) + x²

Solution :

(dy/dx)² + x  =  (dx/dy) + x²

(dy/dx)² + x  =  [1/(dy/dx)] + x²

(dy/dx)³ + x(dy/dx)  =  =  1 + x²(dy/dx)

So, the order is 1 and degree is 3.

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