Question

Given the second-order ordinary differential equation: \( y'' + 3y' - 4y = 0 \) with the initial conditions \( y(0) = 3 \), and \( y'(0) = -7 \), the value of \( y(1) \) is \(\underline{\hspace{2cm}}\) (round off to two decimal places).

Hint:

To solve second-order differential equations, use the characteristic equation to find the general solution and apply initial conditions to find the constants.

Answer 1

Correct Answer: 2.71

The given differential equation is:
\[ y'' + 3y' - 4y = 0 \] We solve the characteristic equation:
\[ r^2 + 3r - 4 = 0 \] Solving for \( r \) using the quadratic formula:
\[ r = \frac{-3 \pm \sqrt{3^2 - 4(1)(-4)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \] Thus, the roots are \( r = 1 \) and \( r = -4 \). The general solution is:
\[ y(x) = C_1 e^x + C_2 e^{-4x} \] Using the initial conditions to find \( C_1 \) and \( C_2 \):
\( y(0) = 3 \) gives:
\[ C_1 + C_2 = 3 \] \( y'(0) = -7 \) gives:
\[ C_1 - 4C_2 = -7 \] Solving the system of equations:
From \( C_1 + C_2 = 3 \), we get \( C_1 = 3 - C_2 \). Substituting into the second equation:
\[ (3 - C_2) - 4C_2 = -7 \implies 3 - 5C_2 = -7 \implies -5C_2 = -10 \implies C_2 = 2 \] Thus, \( C_1 = 3 - 2 = 1 \). The solution is:
\[ y(x) = e^x + 2e^{-4x} \] Now, evaluate \( y(1) \):
\[ y(1) = e^1 + 2e^{-4} \approx 2.718 + 2(0.0183) = 2.718 + 0.0366 = 2.7546 \] Thus, \( y(1) \approx 2.75 \). The value of \( y(1) \) is \( \boxed{2.75} \).