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# 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 = d^{2}y / dx^{2} = r^{2}.e^{rx}

Substituting values of yꞌ and yꞌꞌ in (1),

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

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

Divide (2) by e^{rx}

⇒ 5r^{2} + 14r - 3 = 0

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

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

r = [-14 ± 16] / 10

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

r = 1/5 and -3

## 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.

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