Q2E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{x}$

Step-by-Step Solution

Answer

The particular solution is $y_p=\frac{x^2}{2}+2x$

Step 1: Definition

Variation of parameters, also known as variation of constants, is a general method to solve inhomogeneous linear ordinary differential equations.

Step 2: Find complementary solution

It is given that $y^{\text{'}\text{'}\text{'}}-2y^{\text{'}\text{'}}+y^{\text{'}}=x$

The auxiliary equation is gives:

$$\begin{gathered} m^3-2m^2+m=0 \\ m=0,1,1 \end{gathered}$$

So, the complementary solution is given by $CF=c_1+c_2{e}^x+c_{3}x{e}^x$

Step 3: Calculate Wornkians

Compare with. $CF=c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)+c_3{y}_3\left(x\right)$

$y_1\left(x\right)=1,y_2\left(x\right)=e^x\text{ and }{y}_3\left(x\right)=x{e}^x$

$$\begin{gathered} W\left[\begin{array}{ccc}1& e^x& x{e}^x\end{array}\right]=\left|\begin{array}{c}1\\ 0\\ 0\end{array}\begin{array}{c}{e}^x\\ e^x\\ e^x\end{array}\begin{array}{c}x{e}^x\\ e^x\left(x+1\right)\\ e^x\left(x+2\right)\end{array}\right| \\ =e^{2x} \\ {W}_1=(-1{)}^{3-1}W\left[\begin{array}{cc}x& x{e}^x\end{array}\right] \\ \left|\begin{array}{cc}{e}^x& x{e}^x\\ 1·e^x& e^x\left(x+1\right)\end{array}\right| \\ =e^{2x} \\ {W}_2=(-1{)}^{3-2}W\left[\begin{array}{cc}1& x{e}^x\end{array}\right] \\ -\left|\begin{array}{cc}{e}^x& x{e}^x\\ 0& e^x\left(x+1\right)\end{array}\right| \\ =-e^x(x+1) \\ {W}_3=(-1{)}^{3-3}W\left[\begin{array}{cc}1& e^x\end{array}\right] \\ =\left|\begin{array}{cc}1& e^x\\ 0& e^x\end{array}\right| \\ =e^x \end{gathered}$$

Step 4: Particular solution

The particular solution is given by $y_p\left(x\right)=y_1\left(x\right)·v_1\left(x\right)+y_2\left(x\right)·v_2\left(x\right)+y_3\left(x\right)·v_3\left(x\right)$

Here,

$$\begin{gathered} y_p=1\int \frac{e^{2x}x}{e^{2x}}dx+e^x\int \frac{-e^x(x+1)x}{e^{2x}}dx+x{e}^x\int \frac{e^{x}x}{e^{2x}}dx \\ {y}_p=\frac{x^2}{2}+x{e}^x(-x-1)e^{-x}+e^x\int \frac{\left(-x^2-x\right)}{e^x}dx \end{gathered}$$

$$\begin{gathered} \int \frac{\left(-x^2-x\right)}{e^x}dx=\left(-x^2-x\right)\int e^{-x}dx-\int (-2x-1)\int e^{-x}dx \\ =\left(x^2+x\right)e^{-x}+\int (2x+1)\left(\frac{e^{-x}}{-1}\right)dx \\ =\left(x^2+x\right)e^{-x}-\int (1+2x)e^{-x}dx \\ =\left(x^2+3x+3\right)e^{-x} \end{gathered}$$

$$\begin{gathered} y_p=\frac{x^2}{2}+\left(-x^2-x\right)+x^2+3x+3 \\ =\frac{x^2}{2}+2x+3 \\ {y}_p=\frac{x^2}{2}+2x+3 \end{gathered}$$

$y_p=\frac{x^2}{2}+2x$