Question

Solve the initial value problem: d^2x /dt^2 + 2 dy/dx - 3x = 0, X(2pie) = 1, x^.(2pie) = 13

Answer 1

Answer on Question #50739 – Math – Differential Calculus | Equations

Solve the initial value problem: $d^2x/dt^2 + 2 dy/dx - 3x = 0$, $x(2pie) = 1$, $x^2(2pie) = 13$.

Solution

$$x'' + 2x' - 3x = 0, \quad x(2\pi) = 1, \quad x'(2\pi) = 13.$$

Characteristic equation is $r^2 + 2r - 3 = 0$, hence its solutions are

$$r = -3 \text{ or } r = 1.$$

So, $x(t) = c_1 e^{-3t} + c_2 e^t$, hence $x'(t) = -3c_1 e^{-3t} + c_2 e^t$.

From conditions

$$x(2\pi) = 1, \quad x'(2\pi) = 13$$

obtain the following system of linear equations with respect to $c_1$ and $c_2$, which to be solved by Cramer's method:

\begin{array}{l} \rightarrow \left\{\begin{array}{c}c_1 e^{-6\pi} + c_2 e^{2\pi} = 1 \\ -3c_1 e^{-6\pi} + c_2 e^{2\pi} = 13 \end{array}\right. \rightarrow \left\{ \begin{array}{l} c_1 = \dfrac{\left|\begin{array}{cc}1 & e^{2\pi}\\ 13 & e^{2\pi}\end{array}\right|}{\left|\begin{array}{cc}e^{-6\pi} & e^{2\pi}\\ -3e^{-6\pi} & e^{2\pi}\end{array}\right|} \\ c_2 = \dfrac{\left|\begin{array}{cc}e^{-6\pi} & 1\end{array}\right|}{\left|\begin{array}{cc}-3e^{-6\pi} & 13\end{array}\right|} \\ c_2 = \dfrac{\left|\begin{array}{cc}e^{-6\pi} & e^{2\pi}\end{array}\right|}{e^{2\pi}} \right\} \\ \end{array} \right. \end{array}

Thus, $x(t) = -3e^{6\pi - 3t} + 4e^{t - 2\pi}$ is the solution to the initial value problem.

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