Chapter 4: Q23E (page 164)

In Problems 22–25, use the method described in Problem 21 to find a general solution to the given equation.
23. 5Y' + 4Y = 0

Short Answer

Expert verified

The general solution is $y (t) = c e^{- \frac{4}{5} t}$ .

Step by step solution

Substitute y = ert

The given differential equation is 5y' + 4y = 0

Substitute y = e^rt and y' = re^rt,

$\begin{array}{rcl} 5 (r e^{r t}) + 4 (e^{r t}) & = & 0 \\ (5 r + 4) (e^{r t}) & = & 0 \\ 5 r + 4 & = & 0 \end{array}$

Therefore,

$r = - \frac{4}{5}$

Find the general solution.

Thus, the general solution is $y (t) = c e^{- \frac{4}{5} t}$ .

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In Problems 22–25, use the method described in Problem 21 to find a general solution to the given equation.
23. 5Y' + 4Y = 0

Short Answer

Step by step solution

Substitute y = ert

Find the general solution.

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In Problems 22–25, use the method described in Problem 21 to find a general solution to the given equation.23. 5Y' + 4Y = 0

Short Answer

Step by step solution

Substitute y = ert

Find the general solution.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Mechanics Maths

Decision Maths

Applied Mathematics

Geometry

Statistics

Study anywhere. Anytime. Across all devices.

In Problems 22–25, use the method described in Problem 21 to find a general solution to the given equation.
23. 5Y' + 4Y = 0