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