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Chapter 8: Problem 19

$$9 x^{2} y^{\prime \prime}+9 x^{2} y^{\prime}+2 y=0$$

Short Answer

Expert verified
The solution to this equation depends on the roots of the auxiliary/characteristic equation. Therefore, the general solution of the given equation comes in three forms based on type of roots - real and unequal, real and equal or complex.

Step by step solution

01

Simplify the Equation

First, notice that each term in the equation can be divided by \(9x^{2}\), which gives the simplified equation: \(y'' + y' + \frac{2y} {9x^{2}} = 0\)
02

Set up Auxiliary/Characteristic Equation

Next, for the homogeneous linear differential equation with constant coefficients, an auxiliary or characteristic equation can be formed. This is a quadratic equation in terms of 'm' and corresponds to each derivative with y, y', y'' as m^0, m^1, m^2 respectively, giving us the equation: \(m^{2} + m + \frac{2}{9x^{2}} = 0\)
03

Find the Roots of the Characteristic Equation

To find the roots of this quadratic equation, we can use the Quadratic Formula, which is: \(m = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) After plugging the coefficients into the formula, we can get the value of 'm'.
04

Write the General Solution

Finally, based on the roots of the characteristic equation, we can write down the general solution for the given differential equation. If roots are real and unequal (m1 and m2), the solution is: \( y = C_{1}e^{m_{1}x} + C_{2}e^{m_{2}x}\) If roots are real and equal (m1 =m2), the solution is: \( y = (C_{1} + C_{2}x) e^{m_{1}x} \) If roots are complex (a \pm bi), the solution is: \( y = e^{ax} [C_{1}cos(bx) + C_{2}sin(bx)] \)

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Most popular questions from this chapter

In Problems \(1-10,\) classify each singular point (real or complex) of the given equation as regular or irregular. $$x^{3}(x-1) y^{n}+\left(x^{2}-3 x\right)(\sin x) y^{\prime}-x y=0$$

\(F\left(\frac{1}{2}, 1 ; \frac{3}{2} ;-x^{2}\right)=x^{-1} \arctan x\)

$$\theta^{3} y^{\prime \prime}+\theta(\sin \theta) y^{\prime}-(\tan \theta) y=0, \quad \text { at } \quad \theta=0$$

$$4 x^{2} y^{\prime \prime}+2 x^{2} y^{\prime}-(x+3) y=0$$

\(f(x)=x^{-1}, \quad x_{0}=1\)
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