Find a function y = y(x) that has all of the following properties:• It satisfies the differential equation y" – 5y' + 6y = 0.• Its graph has a y-intercept of (0, 4).• At the y-intercept, its graph has a horizontal tangent line.

Answered: Find a function y = y(x) that has all…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Concept explainers

Question

Find a function y = y(x) that has all of the following properties:
• It satisfies the differential equation y" – 5y' + 6y = 0.
• Its graph has a y-intercept of (0, 4).
• At the y-intercept, its graph has a horizontal tangent line.

Expert Solution

Step 1

We have to find a function $y = y (x)$ that has the following properties.

1) It satisfies the differential equation $y'' - 5 y' + 6 y = 0$ .

2) Its graph has a y intercept of (0,4).

3) At the y-intercept, its graph has a horizontal tangent line.

Step 2

That is, we have to find the solution of the differential equation $y'' - 5 y' + 6 y = 0$ satisfying the conditions $y (0) = 4, y' (0) = 0$ .

The auxiliary equation of the differential equation $y'' - 5 y' + 6 y = 0$ is $m^{2} - 5 m + 6 = 0$ .

Find the roots of the auxiliary equation $m^{2} - 5 m + 6 = 0$ as shown below.

$m^{2} - 5 m + 6 = 0 m^{2} - 2 m - 3 m + 6 = 0 m (m - 2) - 3 (m - 2) = 0 (m - 3) (m - 2) = 0 m_{1} = 3, m_{2} = 2$

Thus, the roots of the auxiliary equation are $m_{1} = 3 and m_{2} = 2$ .

Step 3

Hence the general solution of the differential equation is,

$\begin{array}{rcl} y (x) & = & c_{1} e^{m_{1} x} + c_{2} e^{m_{2} x} \\ = & c_{1} e^{3 x} + c_{2} e^{2 x} \end{array}$

Now evaluate the constants using the initial conditions $y (0) = 4, y' (0) = 0$ as follows.

The derivative of $\begin{array}{rcl} y (x) & = & c_{1} e^{3 x} + c_{2} e^{2 x} \end{array}$ is $\begin{array}{rcl} y' (x) & = & 3 c_{1} e^{3 x} + 2 c_{2} e^{2 x} \end{array}$ .

Then,

$y (0) = 4 c_{1} e^{3 (0)} + c_{2} e^{2 (0)} = 4 c_{1} + c_{2} = 4 \dots \dots (1)$

Trending now

This is a popular solution!

Step by step

Solved in 5 steps

Knowledge Booster

$Differential Equation$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions