Use power series to show that y=e^(x^2) ...

[tyronne]

Use power series to show that y=e^(x^2) is a solution to the differential equation y'-2xy=0.
Help please?

 

[galactus]

Solve your DE by separating variables.

\(\displaystyle y'=2xy\)

\(\displaystyle \frac{y'}{y}=2x\)

\(\displaystyle \int\frac{1}{y}dy=\int{2x}dx\)

\(\displaystyle ln(y)=x^{2}+C\)

\(\displaystyle y=C_{1}e^{x^{2}}\)

 

[tyronne]

hmm, is there a way that this can be solved through power series though? or is that it?

 

[Deleted member 4993]

Use power series to show that y=e^(x^2) is a solution to the differential equation y'-2xy=0.
Help please?

What is the power series expansion of e^(x).

Using that - find power series expansion of e^(x^2).

Then differentiate that to find y' - and plug all those back to your given DE and show that the DE is satisfied.

 

[Dr. Flim-Flam]

f(x) = e^(x^2), Power series e^x =Sum(x^n/n!), n goes from 0 to infinity, hence e^(x^2) = Sum(x^(2n)/n!) as n goes from 0 to infinity.

Note: e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ... +x^n/n!

e^(x^2) = 1 + (x^2) + x^4/2! + x^6/3! + x^8/4! +... +x^(2n)/n!