Question

Use Euler’s method to approximate$\mathit{y}\mathbf{(}\mathbf{1}\mathbf{.2}\mathbf{)}$, where$\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}$is the solution of the given initial-value problem.

.${\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}\text{'}}}\mathbf{-}\mathbf{2}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{9}$

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

Short answer

The solution of the problem calculated using Euler’s method is $y=c_{1}x+c_2{x}^2$and the approximated value for $y\left(1.2\right)$is $5.9$which is lesser than the actual value $6$.

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,

$x^2{y}^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y=0\text{(1)}$

with initial conditions

$y\left(1\right)=4,y^{\text{'}}\left(1\right)=9$

and we have to approximate $y\left(1.2\right)$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 (a) and (b) in equation (1).

$x^2{u}^{\text{'}}-2xu+2y=0\text{(c)}$

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

$\begin{array}{c}{y}^{\text{'}}=u\\ u^{\text{'}}=\frac{2}{x}u-\frac{2}{x^2}y\end{array}$

Hence the first order differential equations are$y^{\text{'}}=u,u^{\text{'}}=\frac{2}{x}u-\frac{2}{x^2}y$

Find the value of y2,u2.

Using Euler’s method using the step $h=0.1$along with the conditions we can have,

$\left(x_0,y_0\right)=(1,4)\text{,}\left(x_0,u_0\right)=(1,9)$

we can have

$\begin{array}{c}{y}_{n+1}=y_1=y_n+h{u}_n\\ =y_0+h{u}_0\\ =4+0.1\times 9\\ =4+0.9\\ =4.9\end{array}$

and

$\begin{array}{c}{u}_1=u_0+h\left[\frac{2}{x_0}{u}_0-\frac{2}{x_0^2}{y}_0\right]\\ =9+0.1\left[\frac{2}{1}\times 9-\frac{2}{1^2}\times 4\right]\\ =9+0.1(18-8)\\ =9+0.1\times 10\\ =9+1\\ =10\end{array}$

we have

$\begin{array}{c}{y}_2=y_1+h{u}_1\\ =4.9+0.1\times 10\\ =4.9+1\\ =5.9\end{array}$

Hence the assumed value is$y_2=5.9$

Find the general solution for the equation.

Now to solve the second order homogenous differential equation shown in we have to follow the below steps:

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

$y^{\text{'}}=m{x}^{m-1}\text{(1a)}$

Differentiate (1a) with respect to x.

$y^{\text{'}\text{'}}=m(m-1)x^{m-2}\text{(2a)}$

Substitute $y=x^m$,(1a), (2a) in equation (1) .

$\begin{array}{c}{x}^2\left(m(m-1)x^{m-2}\right)-2x\left(m{x}^{m-1}\right)+2x^m=0\\ m(m-1)x^m-2m{x}^m+2x^m=0\\ x^m\left(m^2-m-2m+2\right)=0\\ x^m\left(m^2-3m+2\right)=0\end{array}$

Since $x^m$cannot be equal to zero, we have

$\begin{array}{c}{m}^2-3m+2=0\\ (m-1)(m-2)=0\end{array}$

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

$y=c_{1}x+c_2{x}^2\text{(2)}$

Find the approximated value.

Now we have to apply the initial condition $(x,y)=(1,4)$in equation (2)

$\begin{array}{c}4=c_1\times 1+c_2{1}^2\\ c_1+c_2=4\text{(1b)}\end{array}$

Differentiate (1) with respect to x then we get,

$y^{\text{'}}=c_1+2c_{2}x\text{(3)}$

We have to apply the initial condition $\left(x,y^{\text{'}}\right)=(1,9)$in equation (3)

$\begin{array}{c}9=c_1+2c_2{1}^2\\ c_1+2c_2=9\text{(2b)}\end{array}$

Solve (1a) and (2b) .

$c_1=-1\text{,}{c}_2=5$

Now we have to obtain the actual value of y at$x=1.2$ .

$\begin{array}{c}y\left(1.2\right)=5{\left(1.2\right)}^2-1.2\\ =5\times 1.44-1.2\\ =7.2-1.2\\ =6\end{array}$

This value is greater than the approximated value$5.9$

Hence the value of $y$at $x=1.2$is $6$.