- #1
Benny
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Hello, I am having trouble understanding a question in relation to Euler's method.
Basically, the question goes something like Euler's method is solved to solve the differential equation [tex]\frac{{dy}}{{dx}} = \log _e \left( {4 - x^2 } \right)[/tex], with a step size of 0.05 and initial condition y = 0 when x = 0. Let A be magnitude of the area enclosed by the curve [tex]f\left( x \right) = \log _e \left( {4 - x^2 } \right)[/tex], the coordinate axes and the line x = 1. Why is [tex]y_{20}[/tex] an estimate of A?
Answer: [tex]y_{20} \approx \int\limits_0^{x_{20} } {\log _e \left( {4 - x^2 } \right)} dx = \int\limits_0^1 {\log _e \left( {4 - x^2 } \right)dx} = A[/tex]
I do not understand the answer. As far as I understand, [tex]y_{20}[/tex] is just the value of the antiderivative at x = 1, given initial conditions but the answer does not make use of the initial conditions. I do not see how [tex]y_{20}[/tex] can be considered to be an approximation of A if the initial conditions are not used.
Basically, the question goes something like Euler's method is solved to solve the differential equation [tex]\frac{{dy}}{{dx}} = \log _e \left( {4 - x^2 } \right)[/tex], with a step size of 0.05 and initial condition y = 0 when x = 0. Let A be magnitude of the area enclosed by the curve [tex]f\left( x \right) = \log _e \left( {4 - x^2 } \right)[/tex], the coordinate axes and the line x = 1. Why is [tex]y_{20}[/tex] an estimate of A?
Answer: [tex]y_{20} \approx \int\limits_0^{x_{20} } {\log _e \left( {4 - x^2 } \right)} dx = \int\limits_0^1 {\log _e \left( {4 - x^2 } \right)dx} = A[/tex]
I do not understand the answer. As far as I understand, [tex]y_{20}[/tex] is just the value of the antiderivative at x = 1, given initial conditions but the answer does not make use of the initial conditions. I do not see how [tex]y_{20}[/tex] can be considered to be an approximation of A if the initial conditions are not used.