Answer to Question #122449 in Differential Equations for JSE

Question

1. Find all values of m so that the function y = e^mx is a solution of the given differential equation. (Enter your answers as a comma-separated list.)

y prime + 3y = 0

m =____

2. Determine whether Theorem 1.2.1 guarantees that the differential equation
y prime = square root y^2 -25 possesses a unique solution through the given point.
(1, 7)

A. Yes
B. No

Answer 1

1 $y = e^{mx}$ is solution of $\frac{dy}{dx} = 3y$

So differentiate y with respect to x,

$\frac{dy}{dx} = me^{mx}$

now, as per equation, $(m+3)e^{mx} = 0$

solution for this is $m=-3$ only.

2 Given differential equation,

$\frac{dy}{dx} = \sqrt {y^2 - 25}$

Solving this equation,

$\int \frac{dy}{\sqrt {y^2-25}} = \int dx$

$ln|y+\sqrt {y^2-25}| = x + lnC$ . . . . . . . . (i)

At point (1,7), value of constant C,

$ln|7+\sqrt{49-25}| = 1 + lnC$

$lnC = ln|7+\sqrt{24}| -1$

putting back value in equation (i)

$ln|y+\sqrt{y^2 - 25}| = x + ln|7+\sqrt{24}|-1$

This equation is unique solution for differential equation.