Question

For the following problems, verify the given solution and then, by method (e) above, find a second solution of the given equation

$3x{y}^{\text{'}\text{'}}-2(3x-1)y^{\text{'}}+(3x-2)y=0$

Short answer

The proof that of the solution of the differential is stated below and the second solution is $y=\left(e^x{x}^{1/3}\right)$.

Step-by-step solution

Given information

The given differential equation is $3x{y}^{\text{'}\text{'}}-2(3x-1)y^{\text{'}}+(3x-2)y=0$.

Differential equation

A differential equation is an equation that relates one or more unknown functions and their derivatives.

Solution of the equation

Consider the differential equation.

$3x{y}^{\text{'}\text{'}}-2(3x-1)y^{\text{'}}+(3x-2)y=0$

The solution of the equation is,

$u=e^x$

So,

$$\begin{gathered} y=A{e}^x \\ {y}^{\text{'}}=A{e}^x \\ {y}^{\text{'}\text{'}}=A{e}^x \end{gathered}$$

Consider left side of given equation

Consider the left-hand side of the given differential equation.

$$\begin{gathered} 3x{y}^{\text{'}\text{'}}-2(3x-1)y^{\text{'}}+(3x-2)y \\ =3x\left(A{e}^x\right)-2(3x-1)A{e}^x+(3x-2)A{e}^x \\ =A{e}^x(3x-6x+2+3x-2) \\ =A{e}^x\left(0\right) \\ =0 \end{gathered}$$

So,$u=e^x$is the solution of the given differential equation.

Let,

$$\begin{gathered} y=uv \\ =e^{x}v \end{gathered}$$

SO,

$$\begin{gathered} y^{\text{'}}=e^x{v}^{\text{'}}+v{e}^x \\ {y}^{\text{'}\text{'}}=e^x{v}^{\text{'}\text{'}}+e^x{v}^{\text{'}}+e^x{v}^{\text{'}}+v{e}^x \\ =e^x{v}^{\text{'}\text{'}}+2e^x{v}^{\text{'}}+v{e}^x \end{gathered}$$

Equation can be written as

The differential equation can now be written as,

$$\begin{gathered} 3x\left(e^x{v}^{\text{'}\text{'}}+2e^x{v}^{\text{'}}+v{e}^x\right)-2(3x-1)\left(x^x{v}^{\text{'}}+v{e}^x\right)+(3x-2)e^{x}v=0 \\ 3x{e}^x{v}^{\text{'}\text{'}}+2e^x{v}^{\text{'}}=0 \\ \frac{v^{\text{'}\text{'}}}{v^{\text{'}}}=\frac{-2}{3x} \\ \frac{d\left(v^{\text{'}}\right)}{v^{\text{'}}}=-\frac{2}{3x} \end{gathered}$$

Integrate the above equation.

$$\begin{gathered} \ln v^{\text{'}}=\frac{-2}{3}\ln x+\ln K \\ {v}^{\text{'}}=\frac{K}{\left(x^{2/3}\right)} \\ \frac{dv}{dx}=\frac{K}{\left(x^{2/3}\right)} \\ dv=\frac{K}{\left(x^{2/3}\right)}dx \end{gathered}$$

Solve further

Integrate both side of the equation.

$$\begin{gathered} v=\frac{K{x}^{1/3}}{\frac{1}{3}}+C \\ =3K{x}^{1/3}+C \end{gathered}$$

Consider$C=0$.

$v=3K\left(x^{1/3}\right)$

Thus,

$$\begin{gathered} y=uv \\ =3K{e}^x\left(x^{1/3}\right) \\ =3K\left(e^x{x}^{1/3}\right) \end{gathered}$$

So, the second solution of the above equation is,

$y=\left(e^x{x}^{1/3}\right)\text{.}$

Therefore, the proof that of the solution of the differential is stated above and the second solution is$y=\left(e^x{x}^{1/3}\right)$ .