Question

Identify the singular point. Find the general solution that is valid for values of \(t\) on either side of the singular point. $$ t^{2} y^{\prime \prime}+3 t y^{\prime}+y=0 $$

Short answer

Answer: The general solution of the given ODE is \(y(t)=c_1t^{-1}+c_2t^{-1}\ln{t}\), which is valid for values of \(t\) on either side of the singular point \(t=0\).

Step-by-step solution

Identify the singular point

To find the singular point, we need to determine when the coefficient of \(y^{\prime \prime}\) equals to zero, which is \(t^2=0\). Thus, the singular point is \(t=0\).

Change the form of the given ODE

We have given the ODE in the form of Cauchy-Euler equation: $$ t^{2} y^{\prime \prime}+3 t y^{\prime}+y=0. $$ To solve the equation, we assume a solution in the form \(y = t^r\).

Find derivatives and substitute into the ODE

Assuming a solution in the form \(y = t^r\), we will find the derivatives \(y'\) and \(y''\) and then substitute them back into the ODE. The first derivative: $$y^{\prime} = rt^{r-1}$$ The second derivative: $$y^{\prime \prime} = r(r-1)t^{r-2}$$ Substituting the derivatives into the ODE: $$t^2(r(r-1)t^{r-2})+3t(rt^{r-1}) + t^r=0$$

Simplify the equation and find the characteristic equation

By simplifying the equation and collecting the terms with the same power of t, we will obtain the characteristic equation as follows. $$t^{r}[r(r-1)+3r+1]=0$$ Now, we have the characteristic equation: $$r^2 + 2r +1 = 0$$

Solve the characteristic equation for r

We will now solve the characteristic equation to get the values of r. The characteristic equation is a quadratic equation: $$r^2 + 2r +1 = 0$$ Applying the quadratic formula: $$r = \frac{-2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 1}}{2}$$ Simplifying: $$r=\frac{-2\pm\sqrt{0}}{2}=-1$$ This equation has a repeated root of r=-1.

Write down the general solution

Since we have a repeated root of r=-1, we write the general solution as follows: $$y(t)=c_1t^{-1}+c_2t^{-1}\ln{t}$$ This general solution is valid for values of \(t\) on either side of the singular point \(t=0\).

Key concepts

Singular Point

When studying differential equations, particularly those of the Cauchy-Euler type, one important feature to identify is the singular point. In the context of such an equation, a singular point occurs where the coefficient of the highest derivative becomes zero. It's crucial to recognize singular points as they indicate values for which the standard solution approach may not apply, and special methods might be needed.

For the given equation, \[t^{2} y^{\textprime \textprime}+3 t y^{\textprime}+y=0\],we identify the singular point by setting the coefficient of the highest derivative to zero, which means solving for \(t^2=0\).This simplifies down to \(t=0\),indicating the singular point. Understanding where this occurs is essential because it helps us apply the correct solution method for finding the general solution that holds true on either side of this point.

General Solution of Differential Equation

When faced with a differential equation, one fundamental goal is finding its general solution. This solution incorporates arbitrary constants and represents the infinite set of possible specific solutions that satisfy the equation. To find the general solution, especially when dealing with a Cauchy-Euler equation, one typically substitutes a power function of the independent variable, hoping that the resulting algebraic equation (the characteristic equation) will yield roots that guide the construction of the solution.

In our example, after identifying the singular point, we seek a solution of the form: \(y = t^r\).By finding the derivatives \(y^\textprime = rt^{r-1}\) and \(y^{\textprime \textprime} = r(r-1)t^{r-2}\),we can substitute these back into our differential equation to transform it into an algebraic equation. Simplification then leads to our precious characteristic equation, the roots of which provide the exponents for our solution terms. If we find distinct roots, the general solution is a linear combination of those power functions. However, if there are repeated roots, the solution includes logarithmic terms to account for the multiplicity of the root. This way, we derive a complete formula that can apply across different values of the independent variable, excluding the singular point.

Characteristic Equation

The characteristic equation plays a pivotal role in solving second-order linear differential equations, like the Cauchy-Euler equation we're examining. After converting the original differential equation into an algebraic form by substituting assumed solutions and their derivatives, we arrive at this valuable equation.

For our exercise, the assumed solution and its derivatives are plugged into the differential equation and simplified to yield: \[r^2 + 2r +1 = 0\].This is the characteristic equation of our Cauchy-Euler equation. Solving it by finding the roots gives us critical information about the nature of the general solution of the differential equation. In this instance, we see that the characteristic equation is quadratic and has a repeated root of \(r=-1\).This repeated root changes the form of the general solution, introducing a logarithmic component to accommodate the lack of two distinct solutions. Thus, the characteristic equation essentially determines the 'skeleton' of our final solution, onto which we add 'flesh' in the form of arbitrary constants, completing the general solution.