find an and f(x) so that ∑n=0∞ann!xn satisfy f′(x)−f(x)=x2 and f(0)=1here I tried to find f′(x)=∑n=0∞an(n−1)!xn−1 from df(x)dx−f(x)=x2 using first order differential equation I got y=x2+2x+1→ f(x)?but I don't know what is the relation of the summation and first order differential equation. and how to find f(x) and ancan someone give me hint? thanks!

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Jupellodseple804

2022-02-17

find

a_{n}

and

f (x)

so that

\sum_{n = 0}^{\infty} \frac{a_{n}}{n!} x^{n}

satisfy

f^{'} (x) - f (x) = x^{2}

and

f (0) = 1

here I tried to find

f^{'} (x) = \sum_{n = 0}^{\infty} \frac{a_{n}}{(n - 1)!} x^{n - 1}

from

\frac{d f (x)}{d x} - f (x) = x^{2}

using first order differential equation I got

y = x^{2} + 2 x + 1

\to f (x) ?

but I don't know what is the relation of the summation and first order differential equation. and how to find

f (x)

and

a_{n}

can someone give me hint? thanks!

Alissia Head

Beginner2022-02-18Added 5 answers

Hint: If we assume a Taylor expansion of

f (x)

x_{0} = 0

f (x) = \sum_{n = 0}^{\infty} \frac{a_{n}}{n!} x^{n}

what can we conclude for

f (x = 0)

from the Taylor series for the constant

a_{0}

? Write the first terms of the series to see what happens.
The derivative is given as (note the change of the summation index!)

f^{'} (x) = \sum_{n = 1}^{\infty} \frac{a_{n}}{(n - 1)!} x^{n - 1} = \sum_{n = 0}^{\infty} \frac{a_{n} + 1}{n!} x^{n}

.
Plug this into the differential equation:

f^{'} (x) - f (x) = \sum_{n = 0}^{\infty} \frac{a_{n} + 1}{n!} x^{n} - \sum_{n = 0}^{\infty} \frac{a_{n}}{n!} x^{n} = \sum_{n = 0}^{\infty} [\frac{a_{n} + 1}{n!} - \frac{a_{n}}{n!} x^{n} = 0 x^{0} + 0 x^{1} + 1 x^{2} + 0 x^{3} + \dots

What can you conclude by comparing the coefficients? Use the value of

a_{0}

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