Respuesta :

LammettHash
Not clear if you're supposed to find a series solution to the ODE, or solve exactly then give the requested terms of the series expansion of the that solution...

The exact solution is easy enough to find. We can separate the variables:

[tex]y'-xy=x\iff \dfrac{\mathrm dy}{\mathrm dx}=x(1+y)\implies \dfrac{\mathrm dy}{1+y}=x\,\mathrm dx[/tex]
[tex]\implies\ln(1+y)=\dfrac{x^2}2+C[/tex]
[tex]\implies y=Ce^{x^2/2}-1[/tex]

Given that [tex]y(0)=1[/tex], we have

[tex]1=C-1\implies C=2[/tex]

Then recall that

[tex]e^x=\displaystyle\sum_{n\ge0}\frac{x^n}{n!}[/tex]

to write the solution as

[tex]y=2\displaystyle\sum_{n\ge0}\frac{\left(\frac{x^2}2\right)^n}{n!}-1[/tex]
[tex]y=1+2\displaystyle\sum_{n\ge1}\frac{\left(\frac{x^2}2\right)^n}{n!}[/tex]

so the first four terms of the series are

[tex]1+x^2+\dfrac{x^4}4+\dfrac{x^6}{24}[/tex]

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