Solve differential equation using power series

[kahwawashay1]

Find y(x) as a power series satisfying:
y'-2xy=0 , y(0) = 1

attempt:

y=$\sum$$^{∞}_{n=0}$ a_n(x-x₀)ⁿ

y'=$\sum$$^{∞}_{n=0}$ (n+1)a_n+1(x-x₀)ⁿ

y' - 2xy = y' - 2y(x-x₀) - 2x₀y

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

a₁-2x₀a₀ + $\sum$$^{∞}_{n=1}$(x-x₀)ⁿ[(n+1)a_n+1 - 2a_n-1 - 2x₀a_n] = 0

so when n=0 and n=1 we get:
a₁=2x₀a₀

when n>1,
a_n = $\frac{2}{n}$(x₀a_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 x₀ = 0 ?



Also, another problem is that y' - 2xy = x³e^x2, y(0)=1
In this case, any hint as to what should I do with the e^x^2?

 

[Mark44]

Find y(x) as a power series satisfying:
y'-2xy=0 , y(0) = 1

attempt:

y=$\sum$$^{∞}_{n=0}$ a_n(x-x₀)ⁿ

y'=$\sum$$^{∞}_{n=0}$ (n+1)a_n+1(x-x₀)ⁿ

y' - 2xy = y' - 2y(x-x₀) - 2x₀y

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

a₁-2x₀a₀ + $\sum$$^{∞}_{n=1}$(x-x₀)ⁿ[(n+1)a_n+1 - 2a_n-1 - 2x₀a_n] = 0

so when n=0 and n=1 we get:
a₁=2x₀a₀

when n>1,
a_n = $\frac{2}{n}$(x₀a_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 x₀ = 0 ?

This means that y = 1 when x = 0.

Also, another problem is that y' - 2xy = x³e^x2, y(0)=1
In this case, any hint as to what should I do with the e^x^2?

 

[kahwawashay1]

This means that y = 1 when x = 0.

yes I know that but what I meant was, how can I use that to solve the problem? If I already have the recursive formula giving me the coefficients in terms of a_n, of what use is that y(0)=1? On my homework it says y(0)=1 but when my professor was talking about it, he wrote y(x₀)=1, but they are not the same thing?? Is there some relation between them I am not seeing?

I mean if x=0, then
y(0)=$\sum^{∞}_{n=1}$(0-x₀)ⁿ=1
so:
1-x₀+x₀²-x₀³+... = 1
so this implies x₀=0 ? but then all of my coefficients for odd n's become 0..idk if that's supposed to happen...

 

[jackmell]

Also, another problem is that y' - 2xy = x³e^x2, y(0)=1
In this case, any hint as to what should I do with the e^x^2?

Nothing wrong with the odd ones being zero if they have to. And as far as that $e^{x^2}$, just expand it out in it's power series form and equate coefficients on both sides just like you did above except in this case, all the coefficients on the right side are no longer zero.

 

[kahwawashay1]

Nothing wrong with the odd ones being zero if they have to. And as far as that $e^{x^2}$, just expand it out in it's power series form and equate coefficients on both sides just like you did above except in this case, all the coefficients on the right side are no longer zero.

So but then y(0)=1 means x₀=0 and that a₀ =1, right?

 

[Mark44]

Yes, x₀ = 0, so your series should be expanded in powers of x, meaning that your series solutions is y = a₀ + a₁x + a₂x² + ... The initial condition y(0) = 1 determines a₀.

 

[jackmell]

So but then y(0)=1 means x₀=0 and that a₀ =1, right?

That is correct. Also, compute four or so of the coefficients and then compare them to that series you wrote for the right side of the other one.