Question

Use Euler’s method to approximate$\mathit{y}\mathbf{(}\mathbf{0}\mathbf{.2}\mathbf{)}$ , where $\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}$is the solution of the given initial-value problem.

${\mathit{y}}^{\mathbf{\text{'}\text{'}}}\mathbf{-}\mathbf{4}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$.

Use$\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.1}$ . Find the analytic solution of the problem and compare the actual value of $\mathit{y}\mathbf{(}\mathbf{0}\mathbf{.2}\mathbf{)}$with ${\mathit{y}}_{\mathbf{2}}$.

Short answer

The solution of the problem calculated using Euler’s method is $y=c_1{e}^{2x}+c_{2}x{e}^{2x}$and the approximated value for $y\left(0.2\right)$is which is $-1.6800$lesser than the actual value $-1.4918$.

Step-by-step solution

Define Euler’s Method.

Euler's method, which works by resembling a solution curve with line segments, is used to approximate solutions to certain differential equations. The formula used in this method is,

$y_{n+1}=y_n+hf\left(x_n\right)$

where,

is the step size.

Find the first-order differential equations.

We have the second order of differential equation,

$y^{\text{'}\text{'}}-4y^{\text{'}}+4y=0\text{(1)}$

with initial conditions

$y\left(0\right)=-2,y^{\text{'}}\left(0\right)=1$

and we have to approximate using Euler’s method as the following technique.

First we have to assume that,

$y^{\text{'}}=u\text{(a)}$

Then differentiate with respect to we have the following

$y^{\text{'}\text{'}}=u^{\text{'}}\text{(b)}$

Substitute in equation.

$u^{\text{'}}-4u+4y=0\text{(c)}$

Then from $\left(a\right),\left(c\right)$we can have the system of first order differential equations.

role="math" localid="1667982134373" $\begin{array}{c}{y}^{\text{'}}=u\\ u^{\text{'}}=4(u-y)\end{array}$

Hence the first order differential equations are $y^{\text{'}}=u,u^{\text{'}}=4(u-y)$.

Find the value of y2,u2.

Using Euler’s method using the step $h=0.1$along with the conditions $\left(x_0,y_0\right)=(0,-2),\left(x_0,u_0\right)=(0,1)$we can have,

$\begin{array}{c}{y}_{n+1}=y_1=y_n+h{u}_n\\ =y_0+h{u}_0\\ =-2+0.1\times 1\\ =-1.9\end{array}$

and

$\begin{array}{c}{u}_1=u_0+h\left[4\left(u_0-y_0\right)\right]\\ =1+0.1\left[4\right(1-(-2)\left)\right]\\ =1+0.4(1+2)\\ =1+0.4\times 3\\ =1+1.2\\ =2.2\end{array}$

We have,

$\begin{array}{c}{y}_2=y_1+h{u}_1\\ =-1.9+0.1\times 2.2\\ =-1.9+0.22\\ =-1.68\end{array}$

Since $y_1=-1.9\text{,}{u}_1=2.2$we have,

$\begin{array}{c}{u}_2=u_1+h\left[4\left(u_1-y_1\right)\right]\\ =2.2+0.1\left[4\right(2.2-(-1.9)\left)\right]\\ =2.2+0.4(2.2+1.9)\\ =2.2+0.4\times 4.1\\ =2.2+1.64\\ =3.84\end{array}$

Hence the values are $y_2=-1.68\text{,}{u}_2=3.84$.

Find the general solution for the equation.

Now to solve the second order homogenous differential equation shown in $\left(1\right)$we have to follow the below steps:

Assume that $y=e^{mx}$is a solution for this homogenous differential equation. Then differentiate the assumption with respect to $x$. We get,

$y^{\text{'}}=m{e}^{mx}\text{(1a)}$

Differentiate $\left(1a\right)$with respect to.

$y^{\text{'}\text{'}}=m^2{e}^{mx}\text{(2a)}$

Substitute $y=e^{mx},\left(1a\right),\left(2a\right)$,in equation$\left(1\right)$ .

$\begin{array}{c}\left.m^2{e}^{mx}-4m{e}^{mx}\right)+4e^{mx}=0\\ e^{mx}\left(m^2-4m+4\right)=0\end{array}$

Since $e^{mx}$cannot be equal to zero, we have

$\begin{array}{c}{m}^2-4m+4=0\\ {(m-2)}^2=0\end{array}$

Then the eigenvalues are

$m_1=1,m_2=2$

which are real and repeated.

Then the general solution of the homogenous differential equation is

.$y=c_1{e}^{2x}+c_{2}x{e}^{2x}\text{(2)}$

Find the approximated value.

Now we have to apply the initial condition $(x,y)=(0,-2)$in equation$\left(2\right)$

$\begin{array}{c}-2=c_1{e}^0+c_2\left(0\right)e^0\\ c_1=-2\text{(1b)}\end{array}$

Differentiate $\left(1\right)$with respect to x then we get,

$\begin{array}{c}{y}^{\text{'}}=2c_1{e}^{2x}+c_2{e}^{2x}+2c_{2}x{e}^{2x}\\ =2c_1{e}^{2x}+c_2(2x+1)e^{2x}\text{(3)}\end{array}$

We have to apply the initial condition$\left(x,y^{\text{'}}\right)=(0,1)$ in equation (3)

$\begin{array}{c}1=2c_1{e}^0+c_2(0+1)e^0\\ 2c_1+c_2=1\text{(2b)}\end{array}$

Solve (1a) and (2b).

$c_1=-2\text{,}{c}_2=5$

Now we have to obtain the actual value of $y$at $x=0.2$.

$\begin{array}{c}y\left(0.2\right)=5\left(0.2\right)e^{0.4}-2e^{0.4}\\ =1\times 1.4918-2\times 1.4918\\ =-1.4918\end{array}$

This value is greater than the approximated value $-1.6800$.

Hence the value of $y$at $x=0.2$is$-1.4918$ .