Euler's method uses the recurrence relation
[tex]y_{n+1}=y_n+hf(x_n,y_n)[/tex]
to approximate the value of the solution [tex]y(x)[/tex] to the ODE [tex]y'=f(x,y)[/tex].
[tex]xy\dfrac{\mathrm dy}{\mathrm dx}=1+\ln(x^2)\implies y'=\dfrac{1+\ln(x^2)}{xy}=f(x,y)[/tex]
With a step size of [tex]h=0.5[/tex], there will only be two steps necessary to find the approximate value of [tex]y(3)[/tex] based on the initial point [tex]y(2)=5[/tex]. See the attached table below for the computation results.
To demonstrate how the table is generated: Since [tex]y(2)=5[/tex], you are using [tex](x_0,y_0)=(2,5)[/tex].
[tex]y_1=y_0+hf(x_0,y_0)[/tex]
[tex]y_1=5+0.5\times\dfrac{1+\ln(2^2)}{2\times5}[/tex]
[tex]y_1\approx5.1193[/tex]
The next point then uses [tex]x_1=x_0+0.5=2.5[/tex]
[tex]y_2=y_1+hf(x_1,y_1)[/tex]
[tex]y_1=y_1+0.5\times\dfrac{1+\ln(2.5^2)}{2.5y_1}[/tex]
[tex]y_2\approx5.230[/tex]
