Question

Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.

Answer 1

`y = (7)/(10) e^(3(t - 1)) + (3)/(10)e^(-7(t - 1))`

Explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

`m = \frac{-4 +/- \sqrt{4^(2) - 4 X 1 X (-21))} }{2 X 1}\\ = (-4 +/- √(16 + 84) )/(2)\\= (-4 +/- √(100) )/(2)\\= (-4 +/- 10 )/(2)\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7`

So, the solution of the equation is

`y = Ae^{m_(1) t} + Be^{m_(2) t}`

where m₁ = 3 and m₂ = -7.

So,

`y = Ae^(3t) + Be^(-7t)`

Also,

`y' = 3Ae^(3t) - 7e^(-7t)`

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

`y(1) = Ae^(3X1) + Be^(-7X1)\\1 = Ae^(3) + Be^(-7)\\Ae^(3) + Be^(-7) = 1 (1)`

`y'(1) = 3Ae^(3X1) - 7Be^(-7X1)\\0 = 3Ae^(3) - 7Be^(-7)\\3Ae^(3) - 7Be^(-7) = 0 \\3Ae^(3) = 7Be^(-7)\\A = (7)/(3) Be^(-10)`

Substituting A into (1) above, we have

`(7)/(3)B e^(-10)e^(3) + Be^(-7) = 1 \\(7)/(3)B e^(-7) + Be^(-7) = 1\\(10)/(3)B e^(-7) = 1\\B = (3)/(10) e^(7)`

Substituting B into A, we have

`A = (7)/(3) (3)/(10) e^(7)e^(-10)\\A = (7)/(10) e^(-3)`

Substituting A and B into y, we have

`y = Ae^(3t) + Be^(-7t)\\y = (7)/(10) e^(-3)e^(3t) + (3)/(10) e^(7)e^(-7t)\\y = (7)/(10) e^(3(t - 1)) + (3)/(10)e^(-7(t - 1))`

So the solution to the differential equation is

`y = (7)/(10) e^(3(t - 1)) + (3)/(10)e^(-7(t - 1))`