For what values of r does the function y = e^rx satisfy the differential equation 5y'' + 14y' - 3y = 0?

Solution:

Given function is y = e^rx and 

Differential equation 5y'' + 14y' - 3y = 0 -----(1)

Differentiating y w.r.t x, 

yꞌ = dy/dx = r. e^rx

yꞌꞌ = d²y / dx² = r².e^rx

Substituting values of yꞌ and yꞌꞌ in (1),

5y'' + 14y' - 3y = 0

⇒ 5.(r².e^rx ) + 14.(r.e^rx) - 3e^rx = 0 -----(2)

Divide (2) by e^rx

⇒ 5r² + 14r - 3 = 0 

Using the formula to find roots of quadratic equation of standard form ax² + bx + c = 0 ⇒ [-b ± √(b² - 4ac)] / 2a we get,

r = {(-14) ± √[14² - (4 × 5 × (-3))]} / (2 × 5)

r = [-14 ± 16] / 10

⇒ r = (-14 + 16) / 10 and r = (-14 - 16) / 10

r = 1/5 and -3

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For what values of r does the function y = e^rx satisfy the differential equation 5y'' + 14y' - 3y = 0?

Summary:

The function y = e^rx satisfies the differential equation 5y'' + 14y' - 3y = 0 for the values r = 1/5 and -3.