Question

Solve the initial value problem:
(d^2x/dt^2)-3(dx/dt)-10x=0
x(0)=1 and x'(0)=0

Answer 1

Answer on Question#39518 – Math – Other

Solve the initial value problem: $(d^2x/dt^2)-3(dx/dt)-10x=0$, $x(0)=1$ and $x'(0)=0$.

Solution:

Rewrite our equation as $\ddot{x} - 3\dot{x} - 10x = 0$, where $x = x(t)$ be function with the argument $t$. We have linear homogeneous differential equation of second order. The respective characteristic equation $k^2 - 3k - 10 = 0$.

Discriminant of this quadratic equation is $D = 9 - 4(-10) = 9 + 40 = 49$, than $k_{1} = \frac{3 + 7}{2} = 5$, $k_{2} = \frac{3 - 7}{2} = -2$ are the roots of quadratic equation.

By the method of Euler, we have the solution of differential equation $x(t) = c_1 e^{5t} + c_2 e^{-2t}$, where $c_1$ and $c_2$ are the constants.

Let find $c_1$ and $c_2$, and solve the initial value problem.

Let $x(0)=1$. Then

$$1 = c_1 e^{0} + c_2 e^{0} = c_1 + c_2.$$

Let $x'(0)=0$. Then

$$\dot{x}(t) = 5 c_1 e^{5t} - 2 c_2 e^{-2t},$$

whence $0 = 5c_1 e^{0} - 2c_2 e^{0}$ and

$$5c_1 = 2c_2.$$

Then, by (1) and (2)

$$c_1 = \frac{2}{15}, c_2 = \frac{1}{3}.$$

Answer: $x(t) = \frac{2}{15} e^{5t} + \frac{1}{3} e^{-2t}$.