Differential equations power series initial value problem

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Apr 27th, 2015

Since y(0) = 1 and y'(0) = 2, the solution of the initial value problem can be written as the following power series: y(x) = 1 + 2x + a2x2 + a3x3 + a4x4 + … .

Differentiate: y '(x) = 2 + 2a2x + 3a3x2 + 4a4x3 +… and y ''(x) = 2a2 + 6a3x + 12a4x2 + … .

Substitute the expansion of the function y(x) and its second derivative into the differential equation:

y ''(x) – xy(x) = 2a2 + 6a3x + 12a4x2 + … – x(1 + 2x + a2x2 + … ) = 1 + x

and compare the coefficients at the same power on both sides of the equation:

2a2 = 1; 6a3 – 1 = 1; 12a4 – 2 = 0; … . Thus, a2 = 1/2; a3 = 1/3; a4 = 1/6; … ; and

y(x) = 1 + 2x + x2/2 + x3/3 + x4/6 + … 


Apr 27th, 2015

Thank you!

Apr 27th, 2015

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