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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