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Citrus2011 [14]
4 years ago
11

Show that every member of the family of functions y = (6 ln(x) + C)/x , x > 0, is a solution of the differential equation x2y

' + xy = 6. (Simplify as much as possible.) y = 6 ln(x) + C x ⇒ y' = 6x · (1/x) − (6 ln(x) +C) x2 LHS = x2y' + xy = x2 · x2 + x · 6 ln(x) + C x = + 6 ln(x) + C = = RHS, so y is a solution of the differential equation. (b) Illustrate part (a) by graphing several members of the family of solutions on a common screen.
Mathematics
1 answer:
expeople1 [14]4 years ago
3 0

Answer:

y = (6ln(x) + C)/x

A

is not a solution to the differential equation

x²y' + xy = x²

as required. This is probably due to an error in typing either the differential equation, or the value of y.

When you've checked the correct values, the same procedure is followed in obtaining the desired result.

Step-by-step explanation:

Suppose y = (6ln(x) + C)/x

is a solution to the differential equation

x²y' + xy = x²

Differentiate y with respect to x

y' = (6/x)/x - (6ln(x) + C)/x²

= (6 - 6ln(x) - C)/x²

Using this value in the differential equation,

x²(6 - 6ln(x) - C)/x² + x(6ln(x) + C)/x

= 6 - 6ln(x) - C + 6ln(x) + C

= 6

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