Q20E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A particular solution and a fundamental solution set are given for a nonhomogeneous equation and its corresponding homogeneous equation. 

(a) Find a general solution to the nonhomogeneous equation. 

(b) Find the solution that satisfies the specified initial condition.

$$\begin{gathered} \mathbf{xy}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{;} \\ \mathbf{y}\left(1\right)\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\left(1\right)\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(1\right)\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{;} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{;}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\left\{1, x, x^3\right\} \end{gathered}$$

Step-by-Step Solution

Answer

a) The value is $y=c_1+c_2{x}^2+c_3{x}^3+x^2$

b) The value is $y=2+x^2-x^3$

(a) Step 1: Firstly, solve for y n

The given equation is, $xy\text{'}\text{'}\text{'} - y\text{'}\text{'} = - 2$

Solve for, y_n

$xy\text{'}\text{'}\text{'} - y\text{'}\text{'} = 0$

Here it is given that, the fundamental solution set for the homogeneous equation is,   $\left\{1, x, x^3\right\}$

Then, the general solution is $y_n=c_1+c_2{x}^2+c_3{x}^3$

Step 2: A general solution to the nonhomogeneous equation.

$$\begin{gathered} y=y_n+y_p \\ y=c_1+c_2{x}^2+c_3{x}^3+x^2 \end{gathered}$$

(b) Step 3:Solve for given initial conditions.

Given initial conditions are, $y\left(1\right)=2, y\text{'}\left(1\right)=-1, y\text{'}\text{'}\left(1\right)=-4;$

Firstly, solve for, $y\left(1\right)=2$

One has, $y=c_1+c_2{x}^2+c_3{x}^3+x^2$

Substitute $y\left(1\right)=2$ in the above equation,

$$\begin{gathered} 2=c_1+c_2{\left(1\right)}^2+c_3{\left(1\right)}^3+{\left(1\right)}^2 \\ {c}_1+c_2+c_3=1 ......\mathbf{\left(}\mathbf{1}\mathbf{\right)} \end{gathered}$$

Step 4: Now, solve for y ' (1)=-1

One has, $y\text{'}=2c_{2}x+3c_3{x}^2+2x$

Substitute $y\text{'}\left(1\right)=-1$ in the above equation,

$$\begin{gathered} -1=2c_2\left(1\right)+3c_3{\left(1\right)}^2+2\left(1\right) \\ -1=2c_2+3c_3+2 \\ 2c_2+3c_3=-3 ......\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

Step 5: Now, solve for , y '' (1)=-4

One has, $y\text{'}\text{'}=2c_2+6c_{3}x+2$

Substitute $y\text{'}\text{'}\left(1\right)=-4$ in the above equation,

$$\begin{gathered} -4=2c_2+6c_3\left(1\right)+2 \\ 2c_2+6c_3=-6 ......\mathbf{\left(}\mathbf{3}\mathbf{\right)} \end{gathered}$$

Step 6: Find the value of c 1 ,c 2 and c 3

Solve the equation (2) and (3),

$$\begin{gathered} 2c_2+3c_3=-3 \\ 2c_2+6c_3=-6 \\ \overline{-3c_3=3} \\ c_3=-1 \end{gathered}$$

Substitute the value of c₃ in the equation (2),

$$\begin{gathered} 2c_2+3c_3=-3 \\ 2c_2+3\left(-1\right)=-3 \\ 2c_2=0 \\ {c}_2=0 \end{gathered}$$

Substitute the value of c₂ and c₃ in the equation (1),

$$\begin{gathered} c_1+c_2+c_3=1 \\ {c}_1+0+-1=1 \\ {c}_1=2 \end{gathered}$$

Step 7: conclusion, the solution that satisfies the specified initial condition

Substitute the value of c₁,c₂ and c₃ in the general solution.

$$\begin{gathered} y=c_1+c_2{x}^2+c_3{x}^3+x^2 \\ y=\left(2\right)+\left(0\right)x^2+\left(-1\right)x^3+x^2 \\ y=2+x^2-x^3 \end{gathered}$$