Question

Question: Use Euler's method to approximate y(0.2), where y(x) is the solution of the initial-value problem ${\mathbf{y}}^{\text{'}\text{'}}\mathbf{-}\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{y}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}{\mathbf{y}}^{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$. First use one step with h = 0.2 and then repeat the calculations using two steps with h = 0.1.

Short answer

So, the required values are $h=0.1\Rightarrow y\left(0.2\right)=3.23$and$h=0.2\Rightarrow y\left(0.2\right)=3.2$

Step-by-step solution

Definition of Euler's method

The Euler method is a first-order method, which means that the local error (error per step) is proportional to the square of the step size and the global error (error at a given time) is proportional to the step size.

Evaluation

Given initial value problem is $y^{\text{'}\text{'}}-(2x+1)y=1,y\left(0\right)=3,y^{\text{'}}\left(0\right)=1$.

Let $y^{\text{'}}=4$then we get

$u^{\text{'}}-(2x+1)y=1\Rightarrow u^{\text{'}}=1+(2x+1)y$

So, we have system

$$\begin{gathered} {\text{y}}^{\text{'}}=\text{u} \\ {\text{u}}^{\text{'}}=1+(2\text{x}+1){\text{y\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}}_0=3,{\text{u}}_0=1 \end{gathered}$$

Find the value

So, from (3) we get

$$\begin{gathered} {\text{y}}_{\text{n}+1}={\text{y}}_{\text{n}}+h{\text{u}}_{\text{n}} \\ {\text{u}}_{\text{n}+1}={\text{u}}_{\text{n}}+\text{h}\left(1+2{\text{x}}_{\text{n}}\right) \end{gathered}$$

Let h = 0.2 we get

${\text{y}}_1={\text{y}}_0+0.2\times {\text{u}}_0=3+0.2\times 1=3.2$

Substitution

Let h = 0.1 we get

$\begin{array}{rcl}{\text{y}}_1& =& {\text{y}}_0+0.1\times {\text{u}}_0=3+0.1\times 1=3.1a\\ & & \\ {\text{u}}_1& =& {\text{u}}_0+0.1\times \left[2\times {\text{x}}_0+1\right]{\text{y}}_0\\ & =& 1+0.1\times [2\times 0+1].3\\ & =& 1.3\\ & & \\ & & \end{array}$

role="math" localid="1663860053819" $\begin{array}{rcl}{y}_2& =& y_1+0.1\times u_1\\ & =& 3.1+0.1\times \left(1.3\right)\\ y_2& =& 3.23\end{array}$

Thus,

$$\begin{gathered} h=0.2\Rightarrow y\left(0.2\right)=3.2 \\ h=0.1\Rightarrow y\left(0.2\right)=3.23 \end{gathered}$$