Chapter 4: Q14E (page 199)

Find a general solution to the given Cauchy-Euler equation for $t > 0$
$t^{2} y'' (t) - 3 ty' (t) + 4 y (t) = 0$

Short Answer

Expert verified

The general solution to the given equation $t^{2} y'' (t) - 3 ty' (t) + 4 y (t) = 0$ is $y (t) = c_{1} t^{2} + c_{2} t^{2} lnt$ .

Step by step solution

Check the associated characteristic equation.

For a differential equation ${at}^{2} y'' (t) + bty' (t) + cy (t) = f (t)$ , we have that the associated characteristic equation is ${ar}^{2} + (b - a) r + c = 0$ .

In our case, the coefficient of $t^{2} y'' (t)$ is $a = 1$ , the one multiplying is $ty' (t)$ is $b = - 3$ and the one multiplying $y (t)$ is $c = 4$ , therefore the associated characteristic equation is:

$\begin{array}{l} r^{2} + (- 3 - 1) r + 4 = 0 \\ r^{2} - 4 r + 4 = 0 \\ {(r - 2)}^{2} = 0 \\ r = 2, 2 \end{array}$

Check the differential equation.

When one has a double root to the associated characteristic equation, linearly independent solutions to the given differential equation are $y_{1} (t) = t^{r}$ and $y_{2} (t) = t^{r} lnt$ .

So, we have $y_{1} (t) = t^{2}$ , $y_{2} (t) = t^{2} lnt$ and hence the general solution to the given equation is $y (t) = c_{1} t^{2} + c_{2} t^{2} lnt$ .

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Find a general solution to the given Cauchy-Euler equation for $t > 0$
$t^{2} y'' (t) - 3 ty' (t) + 4 y (t) = 0$

Short Answer

Step by step solution

Check the associated characteristic equation.

Check the differential equation.

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Find a general solution to the given Cauchy-Euler equation for t>0t2y''(t)-3ty'(t)+4y(t)=0

Short Answer

Step by step solution

Check the associated characteristic equation.

Check the differential equation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

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Logic and Functions

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Study anywhere. Anytime. Across all devices.

Find a general solution to the given Cauchy-Euler equation for $t > 0$
$t^{2} y'' (t) - 3 ty' (t) + 4 y (t) = 0$