Chapter 4: Q11RP (page 231)

Find a general solution to the given differential equation. $t^{2} x'' (t) + 5 x (t) = 0, t > 0$

Short Answer

Expert verified

$x (t) = c_{1} t^{\frac{1}{2}} \cos (\frac{\sqrt{19}}{2} lnt) + c_{2} t^{\frac{1}{2}} \sin (\frac{\sqrt{19}}{2} lnt)$

Step by step solution

Firstly, write the auxiliary equation of the given differential equation

The differential equation is,

$t^{2} x'' (t) + 5 x (t) = 0 . . . . . . (1)$

The auxiliary equation for the above equation,

$x = t^{m} x' = {mt}^{m - 1} x'' = m (m - 1) t^{m - 2}$

From equation (1),

$t^{2} m (m - 1) t^{m - 2} + 5 t^{m} = 0 m (m - 1) t^{m} + 5 t^{m} = 0 m (m - 1) + 5 = 0 m^{2} - m + 5 = 0$

Find roots of the auxiliary equation

Solve the auxiliary equation,

$m^{2} - m + 5 = 0 m = \frac{- (- 1) \pm \sqrt{1 - 20}}{2} m = \frac{1 \pm \sqrt{- 19}}{2} m = \frac{1 \pm i \sqrt{19}}{2}$

The roots of the auxiliary equation are,

$m_{1} = \frac{1 + i \sqrt{19}}{2}, m_{2} = \frac{1 - i \sqrt{19}}{2}$

$x = t^{m} x = e^{m (lnt)}$

The general solution of the given equation is:

$x (t) = c_{1} t^{\frac{1}{2}} \cos (\frac{\sqrt{19}}{2} lnt) + c_{2} t^{\frac{1}{2}} \sin (\frac{\sqrt{19}}{2} lnt)$

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Find a general solution to the given differential equation. $t^{2} x'' (t) + 5 x (t) = 0, t > 0$

Short Answer

Step by step solution

Firstly, write the auxiliary equation of the given differential equation

Find roots of the auxiliary equation

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Find a general solution to the given differential equation. t2x''t+5xt=0, t>0

Short Answer

Step by step solution

Firstly, write the auxiliary equation of the given differential equation

Find 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

Logic and Functions

Pure Maths

Mechanics Maths

Discrete Mathematics

Decision Maths

Study anywhere. Anytime. Across all devices.

Find a general solution to the given differential equation. $t^{2} x'' (t) + 5 x (t) = 0, t > 0$