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.

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{7}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathit{y}\mathbf{=}\mathbf{24}{\mathit{e}}^{{\mathbf{x}}^{\mathbf{2}}}\mathbf{;}\mathbf{} \\ \mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{5}\mathbf{x}}\mathbf{+}\mathbf{6}{\mathit{e}}^{\mathbf{x}}\mathbf{,}\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)} \end{gathered}$$

Short answer

Proved

Step-by-step solution

Define Cauchy – Euler

The Cauchy – Euler equation or Euler equation is simply a standard equation with a variable coefficient. It is sometimes called the equidimensional equation.

Verify first order differential equation

$y\text{'}\text{'}-7y\text{'}+10y=24e^x$

Given$y=c_1{e}^{2x}+c_2{e}^{5x}+6e^x$,

Then consider,

$$\begin{gathered} y_1=e^{2x} \\ {y}_2=e^{5x} \\ {y}_p=6e^x \end{gathered}$$

Differentiate$y_1$

$$\begin{gathered} y_1\text{'}=2e^{2x} \\ {y}_1\text{'}\text{'}=4e^{2x} \end{gathered}$$

In equation$y\text{'}\text{'}-7y\text{'}+10y=0$ ,

$\begin{array}{rcl}4e^{2x}-7\times 2e^{2x}+10\times e^{2x}& =& 0\\ 14e^{2x}-14e^{2x}& =& 0\\ 0& =& 0\end{array}$

From this we get to know$y_1=e^{2x}$ satisfy the general equation.

Verify second order differential equation

Consider$y_2=e^{5x}$,

Differentiate with respect tox

$$\begin{gathered} y_2\text{'}=5e^{5x} \\ {y}_2\text{'}\text{'}=25e^{5x} \end{gathered}$$

Substitute in equation$y\text{'}\text{'}-7y\text{'}+10y=0$,

$\begin{array}{rcl}25e^{5x}-7\times 5e^{5x}+10\times e^{5x}& =& 0\\ 35e^{5x}-35e^{5x}& =& 0\\ 0& =& 0\end{array}$

From this we get to know $y_2=e^{5x}$ satisfy the general equation.

Use Wronskian method

Substitute the values as we found before,

$\begin{array}{rcl}W\left(y_1,y_2\right)& =& \left|\begin{array}{cc}{y}_1& y_2\\ y_1\text{'}& y_2\text{'}\end{array}\right|\\ & =& \left|\begin{array}{cc}{e}^{2x}& e^{5x}\\ 2e^{2x}& 5e^{5x}\end{array}\right|\end{array}$

Using determinant method and solve matrix,

$\begin{array}{rcl}W\left(e^{2x},e^{5x}\right)& =& \left(e^{2x}\right)\times \left(5e^{5x}\right)-\left(e^{5x}\right)\times \left(2e^{2x}\right)\\ & =& 5e^{7x}-2e^{7x}\\ & =& 3e^{7x}\end{array}$

Since the Wronskian is not zero for all values of x , all the functions are linearly independent on the interval $(-\infty ,\infty )$ .

Hence, from the definition of Fundamental Set of Solutions, the functions form a fundamental set of solutions of the differential equation on the interval $(-\infty ,\infty )$ .

Verify yp  differential equation

Consider$y_p=6e^x$

Differentiate with respect to x

$$\begin{gathered} y_p\text{'}=6e^x \\ {y}_p\text{'}\text{'}=6e^x \end{gathered}$$

Substitute in$y\text{'}\text{'}-7y\text{'}+10y=24e^x$,

$\begin{array}{rcl}6e^x-7\times 6e^x+10\times 6e^x& =& 24e^x\\ 6e^x-42e^x+60e^x& =& 24e^x\\ 66e^x-42e^x& =& 24e^x\\ 24e^{2x}& =& 24e^{2x}\end{array}$

From this we get to know $y_p=6e^x$ satisfy the general equation.

Find y

Substitute

$$\begin{gathered} y_1=e^{2x} \\ {y}_2=e^{5x} \\ {y}_p=6e^x \end{gathered}$$

In general equation we get,

The solution is$y=c_1{e}^{2x}+c_2{e}^{5x}+6e^x$

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.