x'' - 2x' + x = 4t, x(0) = 1, x'(0)=1

Introducing \[y_1 = x\] and \[y_2 = x'\]

we have the system

\[y_1 ' = y_2\]

\[y_2 ' = 2y_2 - y_1 + 4t\]

right?

Using Euler's method with step size h=0.1, we get

\[y_{1, n+1} = y_{1, n} + 0.1y_{2,n}\]

\[y_{2,n + 1} + 0.1(2y_{2,n} - y_{1,n} + 4t)\]

It's easy to see that \[y_{1, 1} = 1 + 0.1 \cdot 1 = 1.1\], but what is \[y_{2,1}\]?

Introducing \[y_1 = x\] and \[y_2 = x'\]

we have the system

\[y_1 ' = y_2\]

\[y_2 ' = 2y_2 - y_1 + 4t\]

right?

Using Euler's method with step size h=0.1, we get

\[y_{1, n+1} = y_{1, n} + 0.1y_{2,n}\]

\[y_{2,n + 1} + 0.1(2y_{2,n} - y_{1,n} + 4t)\]

It's easy to see that \[y_{1, 1} = 1 + 0.1 \cdot 1 = 1.1\], but what is \[y_{2,1}\]?

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