Chapter 4: Q8RP (page 231)

Find a general solution to the given differential equation. $25 y'' + 20 y' + 4 y = 0$

Short Answer

Expert verified

The general solution to the given differential equation is $y = c_{1} e^{\frac{- 2}{5} t} + c_{2} {te}^{\frac{- 2}{5} t} .$

Step by step solution

Write the auxiliary equation of the given differential equation

The given differential equation is $25 y'' + 20 y' + 4 y = 0 .$

The auxiliary equation for the above equation $25 m^{2} + 20 m + 4 = 0 .$

Find the roots of the auxiliary equation.

Solve the auxiliary equation,

$\begin{matrix} 25 m^{2} + 20 m + 4 = 0 \\ {(5 m + 2)}^{2} = 0 \end{matrix}$

The roots of the auxiliary equation are $m_{1} = - \frac{2}{5}, & m_{2} = - \frac{2}{5} .$

Thus, the general solution of the given equation isrole="math" $y = c_{1} e^{\frac{- 2}{5} t} + c_{2} {te}^{\frac{- 2}{5} t} .$

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Find a general solution to the given differential equation. $25 y'' + 20 y' + 4 y = 0$

Short Answer

Step by step solution

Write the auxiliary equation of the given differential equation

Find the roots of the auxiliary equation.

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Find a general solution to the given differential equation.25y''+20y'+4y=0

Short Answer

Step by step solution

Write the auxiliary equation of the given differential equation

Find the roots of the auxiliary equation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Pure Maths

Statistics

Probability and Statistics

Applied Mathematics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

Find a general solution to the given differential equation. $25 y'' + 20 y' + 4 y = 0$