Answer to Question #275862 in Differential Equations for mary

Question

1.  y'''-4y'=0                   ans, y=c₁+c₂e^2x+c₃e^-2x
2. y'' + 9y' + 5y = 0 

Answer 1

Q1.

y'''- 4y' = 0

Let us substitute y by e^mx

So y' = me^mx, y'' = m²e^mx and y''' = m³e^mx

Therefore , m³e^mx - 4me^mx = 0

=> e^mx (m³ - 4m) = 0

=> m³ - 4m = 0

This is the auxiliary equation. Solving it we get

m(m² - 4) = 0

=> m(m - 2)(m + 2) = 0

=> m = 0, 2 , -2

So the general solution of the given differential equation is

y = c₁ + c₂e^2x + c₃e^-2x where c₁, c₂ and c₃ are arbitrary constants.

Q2.

y'' + 9y' + 5y = 0

Let us substitute y by e^mx

So y' = me^mx, y'' = m²e^mx

Therefore, m²e^mx + 9me^mx + 5e^mx = 0

=> e^mx(m² + 9m + 5 ) = 0

=> m² + 9m + 5 = 0

This is the auxiliary equation. Solving it we get

m = $\frac{-9 ± \sqrt{9²-4.1.5}}{2.1}$

=> m = $\frac{-9 ± \sqrt{61}}{2}$ = $\frac{-9}{2} ±\frac{ \sqrt{61}}{2}$

So the general solution of the given differential equation is

y = $c_{1} e^{\frac{-9 + \sqrt{61}}{2}x}+c_{2}e^{\frac{-9 - \sqrt{61}}{2}x}$

=> y = $e^{-\frac{9}{2}x}(c_{1}e^{\frac{\sqrt{61}}{2}x}+c_{2}e^{-\frac{\sqrt{61}}{2}x})$ where c₁ and c₂ are arbitrary constants.