Question

In Problems 31 - 34 verify that the given two-parameter family of functions is the general solution of the nonhomogeneous differential equation on the indicated interval.

$$\begin{gathered} \mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{x}\mathbf{;} \\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{x}}^{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{15}}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{6}}\mathbf{x}\mathbf{,}\mathbf{}\left(0,\infty \right) \end{gathered}$$

Short answer

Proved

Step-by-step solution

Verify the function part satisfies the homogeneous differential equation

We have the non-homogeneous second order differential equation

$2x^{2}y\text{'}\text{'}+5xy\text{'}+y=x^2-x$

With a general solution

$y=c_1{x}^{-\frac{1}{2}}+c_2{x}^{-1}+\frac{1}{15}{x}^2-\frac{1}{6}x$

Where the H. part is $c_1{x}^{-\frac{1}{2}}+c_2{x}^{-1}$and the particular part is$\frac{1}{15}{x}^2-\frac{1}{6}x$.

It is the general solution of the given differential equation on the interval$(0,\infty )$as the following technique:

To verify that the functions of the homogeneous part $y_1=x^{-\frac{1}{2}}and{y}_2=x^{-1}$ satisfy the homogeneous differential equation $2x^{2}y\text{'}\text{'}+5xy\text{'}+y=0$as the following:

For the first function $y_1=x^{-\frac{1}{2}}$:

we have to differentiate this function with respect to x, then we have

$y_1\text{'}=-\frac{1}{2}{x}^{-\frac{3}{2}},y_1^{\text{'}\text{'}}=\frac{3}{4}{x}^{-\frac{5}{2}}$

Substitute with these derivatives into the given differential equation, then we have

$\begin{array}{rcl}2x^2\times \frac{3}{4}{x}^{-\frac{5}{2}}+5x\times -\frac{1}{2}{x}^{-\frac{3}{2}}+x^{-\frac{1}{2}}& =& 0\\ \frac{3}{2}{x}^{-\frac{1}{2}}-\frac{5}{2}{x}^{-\frac{1}{2}}+x^{-\frac{1}{2}}& =& 0\\ -x^{-\frac{1}{2}}+x^{-\frac{1}{2}}& =& 0\\ 0& =& 0\end{array}$

Determine the set of function by using Wronskian

For the second function$y_2=x^{-1}$:

we have to differentiate this function with respect to x, then we have

$y_2\text{'}=-x^{-2},y_2\text{'}\text{'}=2x^{-3}$

After that, we have to substitute with these derivatives into the given differential equation, then we have

$\begin{array}{rcl}2x^2\times 2x^{-3}+5x\times -x^{-2}+x^{-1}& =& 0\\ 4x^{-1}-5x^{-1}+x^{-1}& =& 0\\ 5x^{-1}-5x^{-1}& =& 0\\ 0& =& 0\end{array}$

Then the two functions of the homogeneous part satisfy the homogeneous D.E$2x^{2}y\text{'}\text{'}+5xy\text{'}+y=0$.

Second, we have to know if the functions of the homogeneous solution are linearly dependent or independent using the Wronskian as

$$\begin{gathered} W\left(y_1,y_2\right)=\left|\begin{array}{cc}{y}_1& y_2\\ y_1\text{'}& y_2\text{'}\end{array}\right| \\ =\left|\begin{array}{cc}{x}^{-\frac{1}{2}}& x^{-1}\\ -\frac{1}{2}{x}^{-\frac{3}{2}}& -x^{-2}\end{array}\right| \end{gathered}$$

After that, we have to find the determinate of this matrix as

$\begin{array}{rcl}W\left(x^{-\frac{1}{2}},x^{-1}\right)& =& x^{-\frac{1}{2}}\times -x^{-2}-x^{-1}\times -\frac{1}{2}{x}^{-\frac{3}{2}}\\ & =& -x^{-\frac{5}{2}}+\frac{1}{2}{x}^{-\frac{5}{2}}\\ & =& -\frac{1}{2}{x}^{-\frac{5}{2}}\end{array}$

Since the determinate of the Wronskian of the given set of functions is not equal Zero, then this set of function is linearly independent, and the given functions can form a fundamental set of solution.

Verify the function part satisfies the non-homogeneous differential equation:

For the particular function$y_p=\frac{1}{15}{x}^2-\frac{1}{6}x$ satisfy the given non-homogeneous differential equation $2x^{2}y\text{'}\text{'}+5xy\text{'}+y=x^2-x$ as the following:

Differentiate the particular solution with respect to x, then we have

$y_p\text{'}=\frac{2}{15}x-\frac{1}{6}and{y}_p\text{'}\text{'}=\frac{2}{15}$

After that, we have to substitute with these derivatives into the given differential equation as

$\begin{array}{rcl}2x^2\times \frac{2}{15}+5x\times \left(\frac{2}{15}x-\frac{1}{6}\right)+\frac{1}{15}{x}^2-\frac{1}{6}x& =& x^2-x\\ \frac{4}{15}{x}^2+\frac{2}{3}{x}^2-\frac{5}{6}x+\frac{1}{15}{x}^2-\frac{1}{6}x& =& 24e^x\\ \frac{4+10+1}{15}{x}^2-\frac{5+1}{6}x& =& x^2-x\\ x^2-x& =& x^2-x\end{array}$

Then the particular part $y_p=\frac{1}{15}{x}^2-\frac{1}{6}x$ satisfy the given differential equation.

In conclusion, we find that the general solution$y=c_1{x}^{-\frac{1}{2}}+c_2{x}^{-1}+\frac{1}{15}{x}^2-\frac{1}{6}x$ is valid on the indicated interval$(0,\infty )$.

Hence, it is verified that the given two-parameter family of functions is the general solution of the nonhomogeneous differential equation on the indicated interval.