Question

In Problems 1-16 the indicated function ${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ is a solution of the given differential equation. Use reduction of order or formula (5), as instructed, to find a second solution ${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{1}}\mathbf{=}{\mathit{e}}^{\mathbf{2}\mathbf{x}}$

Short answer

The second solution of the given differential equation is $x{e}^{2x}$

Step-by-step solution

Definition of Reduction of order

• Reduction of order is a method for solving second-order linear ordinary differential equations in mathematics.
• When one solution is known and a second linearly independent solution is sought, this method is used.
• The approach can be used to solve n-th order equations as well.
• $y_2=y_1\left(x\right)\int \frac{e^{-\int P\left(x\right)dx}}{y_1^2\left(x\right)}dx$

Find the second solution

The second order differential equation

$y\text{'}\text{'}-4y\text{'}+4y=0$

as the form

$y\text{'}\text{'}+p\left(x\right)y\text{'}+q\left(x\right)y=0,\text{Where}p\left(x\right)=-4\text{andq(x)=4}$

with a first solution$y_1\left(x\right)=e^{2x}$, and we have to obtain the second solution as

$\begin{array}{rcl}{y}_2\left(x\right)& =& y_1\left(x\right)\int \frac{e^{-\int p\left(x\right)dx}}{{\left(y_1\right)}^2}dx\\ & =& e^{2x}\int \frac{e^{-\int -4dx}}{{\left(e^{2x}\right)}^2}dx\\ & =& e^{2x}\int \frac{e^{4x}}{e^{4x}}dx\\ & =& e^{2x}\int 1dx\\ & =& x{e}^{2x}\end{array}$

Therefore, the second solution of the given differential equation is $x{e}^{2x}$