Chapter 4: Q16E (page 164)

In Problems 13–20, solve the given initial value problem.
y"-4y'-5y = 0 : y(-1) = 3,y'(-1) = 9

Short Answer

Expert verified

The solution is $y (t) = - \frac{2}{3} e^{t - 1} + \frac{5}{3} e^{5 + 5 t}$ .

Step by step solution

Find the solution of the differential equation.

The given differential equation is y" - 4y' + 3y = 0.

The auxiliary equation is;

$\begin{array}{rcl} r^{2} - 4 r^{1} - 53 t^{0} & = & 0 \\ r^{2} - 4 r - 5 & = & 0 \\ r^{2} + r - 5 r - 5 & = & 0 \\ r (r + 1) - 5 (r + 1) & = & 0 \\ (r + 1) (r - 5) & = & 0 \\ r & = & - 1, 5 \end{array}$

Therefore, the solution is y(t) = c₁e^-t + c₂e^5t.

Apply initial conditions.

The initial conditions are y(-1) = 3,y'(-1) = 9.

Therefore,

$\begin{array}{rcl} y (- 1) & = & c_{1} e^{1} + c_{2} e^{- 5} \\ c_{1} e^{1} + c_{2} e^{- 5} & = & 3 \end{array}$

And

$\begin{array}{rcl} y' (t) & = & - c_{1} e^{- t} + 5 c_{2} e^{5 t} \\ y' (- 1) & = & - c_{1} e^{1} + 5 c_{2} e^{- 5} \\ - c_{1} e^{1} + 5 c_{2} e^{- 5} & = & 9 \end{array}$

Solving for c₁,c₂ then,

$c_{1} = - \frac{2}{3} e^{- 1} c_{2} = \frac{5}{3} e^{5}$

Therefore, the solution is $y (t) = - \frac{2}{3} e^{t - 1} + \frac{5}{3} e^{5 + 5 t}$ .

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In Problems 13–20, solve the given initial value problem.
y"-4y'-5y = 0 : y(-1) = 3,y'(-1) = 9

Short Answer

Step by step solution

Find the solution of the differential equation.

Apply initial conditions.

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In Problems 13–20, solve the given initial value problem.y"-4y'-5y = 0 : y(-1) = 3,y'(-1) = 9

Short Answer

Step by step solution

Find the solution of the differential equation.

Apply initial conditions.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Statistics

Theoretical and Mathematical Physics

Probability and Statistics

Discrete Mathematics

Calculus

Study anywhere. Anytime. Across all devices.

In Problems 13–20, solve the given initial value problem.
y"-4y'-5y = 0 : y(-1) = 3,y'(-1) = 9