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

Answer:

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

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

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,

$y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1 (1)$

$y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}$

Substituting A into (1) above, we have

$\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1 \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}$

Substituting B into A, we have

$A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}$

Substituting A and B into y, we have

$y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

So the solution to the differential equation is

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$