Find y(x) as a power series satisfying:(adsbygoogle = window.adsbygoogle || []).push({});

y'-2xy=0 , y(0) = 1

attempt:

y=[itex]\sum[/itex][itex]^{∞}_{n=0}[/itex] a_{n}(x-x_{0})^{n}

y'=[itex]\sum[/itex][itex]^{∞}_{n=0}[/itex] (n+1)a_{n+1}(x-x_{0})^{n}

y' - 2xy = y' - 2y(x-x_{0}) - 2x_{0}y

substituting the power series into above formula and simplifying eventually gives:

a_{1}-2x_{0}a_{0}+ [itex]\sum[/itex][itex]^{∞}_{n=1}[/itex](x-x_{0})^{n}[(n+1)a_{n+1}- 2a_{n-1}- 2x_{0}a_{n}] = 0

so when n=0 and n=1 we get:

a_{1}=2x_{0}a_{0}

when n>1,

a_{n}= [itex]\frac{2}{n}[/itex](x_{0}a_{n-1}+ a_{n-2})

but what does my professor mean by y(0) = 1? I think he was trying to explain that this means that x_{0}= 0 ?

Also, another problem is that y' - 2xy = x^{3}e^{x2}, y(0)=1

In this case, any hint as to what should I do with the e^x^2????

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# Homework Help: Solve differential equation using power series

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